Joel paid $138 for 2 pairs of pants and 3 shirts. Doug paid $204 for 3 pairs of pants and 6 shirts. Set up and
solve a system of equations to find the price of one pair of pants.

The value of x in each case:

(a) x = 7 units

(b) x = 5√2 units

(c) x = 4√3 units

In this question we use some basic formula of trigonometry.

(a) Consider sine of angle 30 degrees

sin(30) = opposite side/hypotenuse

1/2 = x/14

x = 14/2

x = 7 units

(b) Consider cosine of angle 45 dgrees

cos(45) = adjacent side/ hypotenuse

1/√2 = 5/x

x = 5√2 units

(c) Consider tangent of angle 60 degrees.

tan(60) = opposite side/ hypotenuse

√3 = x/4

x = 4 × √3

x = 4√3 units

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The volume of the solid obtained by rotating the region bounded by the curves y = x², y = 0, and x = 1 about the y-axis is π/3 cubic units.

To find the volume of the solid obtained by rotating the region bounded by the curves y = x², y = 0, and x = 1 about the y-axis, we can use the disk method.

The idea behind the disk method is to slice the solid into thin disks perpendicular to the axis of rotation and sum up their volumes. The volume of each disk is the product of its cross-sectional area and its thickness.

In this case, we are rotating about the y-axis, so the cross-sectional area of each disk will be a circle with radius x and area πx². The thickness of each disk will be dx, which represents an infinitesimal slice of the x-axis.

Thus, the volume of each disk is given by:

dV = πx² dx

To find the total volume of the solid, we need to integrate this expression over the range of x from 0 to 1:

V = ∫₀¹ πx² dx

Integrating this expression gives:

V = π/3

Therefore, the volume of the solid obtained by rotating the region bounded by the curves y = x², y = 0, and x = 1 about the y-axis is π/3 cubic units.

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The four possible times that Car B could take to complete one lap are 2 seconds, 3 seconds, 5 seconds, and 15 seconds.

150 is represented as a factor of 2, 3, 5 and 5. As 150/t is considered as an integer, t should be a divisor of 150. Hence, we can consider the potential values for t which are:

- t=2 seconds (because 150/2 equals to 75, making it an integer)

- t=3 seconds (as 150/3 equates to 50, also forming an integer)

- t=5 seconds (since 150 divided by 5 produces 30 which is another integer)

- t=15 seconds (seeing that when dividing 150 with 15 gives us 10 which is likewise an integer).

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a) The elasticity of the demand function q = 800 - x is -x / (800 - x)².

b) At x = 3, the elasticity of the demand function q = 800 - x is approximately -0.0000465.

(a) To find the elasticity of the demand function q = 800 - x, we first need to calculate the derivative of q with respect to x:

dq/dx = -1

Next, we can use the formula for elasticity:

E = (dq/dx) * (x/q)

Substituting the values of dq/dx and q, we get:

E = (-1) * (x/(800-x))

Simplifying this expression, we get:

E = -x / (800 - x)²

(b) To find the elasticity at x = 3, we substitute x = 3 into the expression we derived for E:

E = -(3) / (800 - 3)² = -0.0000465

Therefore, the elasticity at x = 3 is approximately -0.0000465.

Note that since the elasticity is negative, this indicates that the demand is inelastic, meaning that a change in price will have a relatively small effect on the quantity demanded.

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The probability that the weight of 36 babies born at the hospital will be between 7.4 and 6.5 pounds is approximately 0.9089 or 90.89%.

a) The values of the parameters are:

Mean (μ) = 7.3 pounds

Standard deviation (σ) = 0.65 pounds

b) As we don't have the sample data, we can't calculate the sample mean (μx) and sample standard deviation (σx or s).

c) To find the probability that the weight of 36 babies will be between 7.4 and 6.5 pounds, we need to use the central limit theorem as the sample size is large enough (n=36).

First, we need to standardize the values using the formula:

z = (x - μ) / (σ / sqrt(n))

where x is the value we want to find the probability for, μ and σ are the population mean and standard deviation respectively, and n is the sample size.

For 7.4 pounds:

z1 = (7.4 - 7.3) / (0.65 / sqrt(36)) = 1.38

For 6.5 pounds:

z2 = (6.5 - 7.3) / (0.65 / sqrt(36)) = -2.46

Next, we need to find the probability of z-values using a standard normal distribution table or calculator.

Using the standard normal distribution table, the probability of z1 = 1.38 is 0.9157, and the probability of z2 = -2.46 is 0.0068.

Finally, we can find the probability that the weight of 36 babies will be between 7.4 and 6.5 pounds by subtracting the probability of z2 from the probability of z1:

P(6.5 ≤ x ≤ 7.4) = P(z2 ≤ z ≤ z1) = P(z ≤ 1.38) - P(z ≤ -2.46) = 0.9157 - 0.0068 = 0.9089

Therefore, the probability that the weight of 36 babies born at the hospital will be between 7.4 and 6.5 pounds is approximately 0.9089 or 90.89%.

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Answer:

[tex]a. \quad \dfrac{\boxed{1}}{\boxed{3}}} \cdot \dfrac{\boxed{1}}{\boxed{2}}} = \dfrac{\boxed{1}}{\boxed{6}}}\\\\\\\\b. \quad \dfrac{\boxed{1}}{\boxed{3}}} \cdot \dfrac{\boxed{1}}{\boxed{3}}} = \dfrac{\boxed{1}}{\boxed{9}}}\\\\\\\\c. \quad \dfrac{\boxed{21}}{\boxed{26}}} \cdot \dfrac{\boxed{5}}{\boxed{26}}} = \dfrac{\boxed{105}}{\boxed{676}}}\\[/tex]

Step-by-step explanation:

a.

When rolling a die(number cube) the sample space which is the set of all possible outcomes is {1, 2, 3, 4, 5, 6}

The probability of getting any single number on the face is the same = 1/6

For the first cube
P(3) = 1/6 and P(4) = 1/6
P(3 or 4) = P(3) + P(4) = 1/6 + 1/6 = 2/6 = 1/3

For the second cube
P(odd) = P(1 or 3 or 5) = P(1) + P(3) + P(5) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2

So the combined probability that the first cube shows 3 or 4 and the second an odd is given by
1/3 · 1/2 = 1/6

b.
There are three coins a penny, dime and quarter

Probability of selecting a penny = number of pennies/total number of coins = 1/3

Since we are replacing the selected coin for the second draw, the probability of selecting a penny is just the same as before = 1/3

P(selecting 2 pennies with replacement) = 1/3 · 1/3 = 1/9

c.

There are a total of 26 letters in the alphabet
There are 5 vowels in the alphabet: A, E, I, O, U

Therefore there are 26 - 5 = 21 consonants

P(drawing a consonant) = 21/26

P(drawing a vowel) = 5/26

Since we are replacing the first drawn letter, these probabilities do not change with successive draws.

Therefore
P(consonant first draw and vowel second draw)

= P(consonant) · P(vowel)

= 21/26 · 5/26

=105/676

The correct answer is: t = 2.293, p-value < 0.01, reject the null hypothesis, and conclude that the mean weight of the citizens has increased.

To test the hypothesis, we need to use a one-sample t-test since we are comparing the mean weight difference of the sample to zero (no change). The sample mean weight difference is 3.51, and the sample standard deviation is 7.44. Since we do not know the population standard deviation, we use the t-distribution.

The null hypothesis is that the mean weight difference is equal to zero (no change), and the alternative hypothesis is that the mean weight difference is greater than zero (increase in weight).

Using a significance level of 0.01, the critical t-value for a one-tailed test with 40 degrees of freedom is 2.704. The calculated t-value is (3.51-0)/(7.44/sqrt(41)) = 2.293. The p-value associated with this t-value is less than 0.01 (found using a t-distribution table or calculator).

Since the p-value is less than the significance level, we reject the null hypothesis and conclude that the mean weight of the citizens of Pawnee significantly increased after drinking the new candy bar milkshake for a week. Therefore, the correct answer is: t = 2.293, p-value < 0.01, reject the null hypothesis, and conclude that the mean weight of the citizens has increased.

The 95% confidence interval for the mean difference in weight (1.163 , 5.857) also supports this conclusion since it does not include zero.

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The value of p3 for which the set s is linearly dependent is p3 = 1

To find the value of p3 for which s is linearly dependent, we need to find values of p3 such that at least one of the vectors in the set s can be written as a linear combination of the other vectors in the set.

Let's start by setting up the linear combination equation:

c1(t − t^2) + c2(t^2 − t^3) + c3(1 − t t^3) = 0

where c1, c2, and c3 are constants that represent the coefficients of the linear combination.

Next, we can expand the terms and group them by powers of t:

(c1 − c2)t^2 + (-c1 + c2 + c3)t + (c3) = 0

For this equation to have a non-trivial solution (i.e., not all c1, c2, and c3 are zero), the determinant of the coefficient matrix must be zero:

det

|1 -1 0|

|-1 1 1|

|0 -1 p3|

= 0

Expanding the determinant, we get:

1(1-p3) - (-1)(-p3) + 0(1) = 0

Simplifying this equation, we get:

p3 - 1 = 0

Therefore, the value of p3 for which the set s is linearly dependent is p3 = 1

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The given differential equation, y'(t) = 4y e, is separable. and Option A. The solution to the initial value problem is y(t) = e^(4e) is the right answer for the given question.

To determine if the equation is separable, we need to check if we can write the equation in form f(y)dy = g(t)dt. If we rearrange the equation, we get y'(t) = 4y(t)e.

We can write this as y'(t)/y(t) = 4e. Now we can see that we have separated the variables y and t on either side of the equation, so the equation is separable.

To solve the equation, we can integrate both sides with respect to t and y. On the left side, we get ln|y(t)|, and on the right side, we get 4et + C, where C is the constant of integration. Therefore, we have ln|y(t)| = 4et + C.

To find the value of C, we use the initial condition y(0) = 1. Substituting t = 0 and y(t) = 1 into the equation, we get,
ln|1| = 4e(0) + C, so C = ln|1| = 0.

Therefore, Option A. The equation is seperable. The solution to the initial value problem is y(t)=e^4e is the correct answer.

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The mean of the variable given in the question is Option A. 41.6.

To find the mean of the variable, we need to multiply each range of weekly hours worked by its corresponding probability, then sum all of the results.

The calculations are as follows:

(23 * 0.08) + (36 * 0.10) + (43 * 0.74) + (54 * 0.08) = 41.6

Therefore, the mean of the variable is Option A. 41.6.

In this case, the probabilities for each range of weekly hours worked to represent the likelihood of an employee working within that range. For example, the probability of an employee working between 41-50 hours is 0.74, which is quite high compared to the other ranges. As a result, this range has a larger impact on the overall mean of the variable.

It is important to calculate the mean of a variable as it helps in understanding the central tendency of a distribution. In this case, the mean helps us to understand the average number of weekly hours worked by employees, which can be useful in making decisions related to employee scheduling, workload management, and compensation.

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Answer: 5,040

Step-by-step explanation: The value of 7P6 is 5,040. To evaluate this, you can use the formula for permutations: nPr = n! / (n - r)!, where n is the total number of items and r is the number of items being selected. In this case, n = 7 and r = 6, so 7P6 = 7! / (7 - 6)! = 7! / 1! = 5,040.

The polygon is a parallelogram and rectangle

The polygon is a parallelogram , quadrilateral and a rectangle

The sum of angles of a parallelogram is 360°

The four types are parallelograms, squares, rectangles, and rhombuses

Opposite sides are parallel

Opposite sides are congruent

Opposite angles are congruent.

Same-Side interior angles (consecutive angles) are supplementary

Each diagonal of a parallelogram separates it into two congruent triangles

The diagonals of a parallelogram bisect each other

Given data ,

The polygon is represented as OPQR

Now , the number of sides of the polygon = 4

So , it is a quadrilateral

Now , the measure of sides of the quadrilateral are

OP = 20 units

PQ = 40 units

QR = 20 units

RO = 40 units

So, it has 2 congruent sides and they are parallel in shape

So, it is a parallelogram

Now, the 2 opposite pairs of sides of the parallelogram are equal

So, it is a rectangle

Hence, the polygon is a parallelogram and rectangle

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The value of the missing angle x is 114 degrees

From the question, we have the following parameters that can be used in our computation:

The kite

The value of x can be calculated using the following equation

x + 78 + 78 + 90 = 360 ---- sum of angles in a quadrilateral

When the like terms are evaluated, we have

x + 246 = 360

So, we have

x = 114

Hence, the value of x is 114 degrees

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Answer:

y= 0 , 2 , 4

Step-by-step explanation:

y=x+2

substitute the value for x

y=-2+2

y=0

y=0+2

y=2

y=2+2

y=4

There might be a typo in the interval you provided, as it should be written in ascending order, such as [a, b] with a < b. I'll assume you meant the interval [2, 14]. Now, let's analyze the function f(x) = 1/ln(3x) and find its absolute maximum and minimum on the interval [2, 14].

Step 1: Find the critical points
To find the critical points, we need to find the derivative of the function f(x) and set it equal to zero.

f(x) = 1/ln(3x)
Using the chain rule, we find the derivative:
f'(x) = -1/(ln(3x))^2 * (1/x)

Now, we need to find when f'(x) = 0 or when f'(x) is undefined. Since the derivative is a fraction, it is never equal to zero. However, the function is undefined when the denominator is zero. In this case, there's no value of x in the interval [2, 14] that makes the denominator zero.

Step 2: Analyze the endpoints
Since there are no critical points within the interval, we only need to check the values of the function at the endpoints.

f(2) = 1/ln(6)
f(14) = 1/ln(42)

Step 3: Determine the absolute maximum and minimum
Compare the values at the endpoints:
f(2) > f(14) as ln(6) < ln(42)

Thus, the function f(x) has an absolute maximum at x = 2 and an absolute minimum at x = 14 within the interval [2, 14].

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The arc length is approximately 20.5744 units.

To find the arc length from (0, 4) clockwise to (3, 7) along the circle[tex]x^2 + y^2 = 16.[/tex]

We need to first find the angle between the positive x-axis and the line connecting the center of the circle to the point (3, 7), since the arc length is a fraction of the circumference of the circle.

The center of the circle is at (0, 0), and the line connecting (0, 0) to (3, 7) has a slope of (7-0)/(3-0) = 7/3.

Therefore, the angle between the positive x-axis and this line is given by:

θ = arctan(7/3) ≈ 1.1659045 radians

Since we are traveling clockwise from (0, 4) to (3, 7), we are traversing an angle of 2π - θ, which is approximately 5.1766375 radians.

The circumference of the circle is given by 2πr, where r is the radius of the circle. In this case, the radius is 4, so the circumference is 8π.

The fraction of the circumference that we travel along is the ratio of the angle we traverse to the total angle around the circle, which is 2π.

Therefore, the arc length is:

(5.1766375 radians / 2π) × 8π = 20.5744

Rounding to four decimal places, the arc length is approximately 20.5744 units.

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a) The 20th percentile is 21.4.

b) The 40th percentile is 31.

c) The 70th percentile is 66.4.

To determine the percentile, we need to first arrange the data in order from smallest to largest

11, 24, 28, 33, 37, 46, 64, 72

a) To find the 20th percentile, we need to first determine the rank of this percentile.

The formula for rank is given by:

Rank = (percentile/100) x (number of observations + 1)

So for the 20th percentile, we have:

Rank = (20/100) x (8+1) = 1.8

This tells us that the 20th percentile lies between the 1st and 2nd observations. To find the actual value, we can use linear interpolation

Value = 11 + 0.8 x (24 - 11) = 11 + 0.8 x 13 = 21.4

Therefore, the 20th percentile is 21.4.

b) To find the 40th percentile, we use the same formula:

Rank = (40/100) x (8+1) = 3.6

This tells us that the 40th percentile lies between the 3rd and 4th observations. Using linear interpolation:

Value = 28 + 0.6 x (33 - 28) = 28 + 0.6 x 5 = 31

Therefore, the 40th percentile is 31.

c) To find the 70th percentile, we use the same formula:

Rank = (70/100) x (8+1) = 6.3

This tells us that the 70th percentile lies between the 6th and 7th observations. Using linear interpolation:

Value = 64 + 0.3 x (72 - 64) = 64 + 0.3 x 8 = 66.4

Therefore, the 70th percentile is 66.4.

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c) The interest due on January 11th is $66 and the balance of the loan after the January 11th payment is $4,263.

d) The final payment that Peter must make on the due date (interest and principal) is $4,285.

The interest due, balances, and final payments are based on compound interest.

The compound interest system charges interest on both the accumulated interest and principal (balance).

April 15th to October 12th = 180 days

October 13th to January 11th = 90 days

January 12th to February 28th = 47 days

Total number of days for the loan = 317 days

Days in the year = 365 days

Principal = $10,000

Loan period = 317 days

Interest rate = 4%

Balance on October 12th = $10,197 ($10,000 + $10,000 x 4% x 180/365)

Payment on October 12th = $3,500

Balance from October 13th = $6,697 ($10,197 - $3,500)

Balance on January 11th = $6,763 ($6,697 + $66)

Interest = $66 ($6,697 x 4% x 90/365)

Payment on January 11th = $2,500

Balance from January 12th = $4,263 ($6,763 - $2,500)

Final Payment on February 28th = $4,285 ($4,263 + $4,263 x 4% x 47/365)

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We can use the Taylor series formula to find the fourth-degree Taylor polynomial for f about x = 2.  The answer is d. 18

[tex]f(2) = f(2) = 405[/tex]

[tex]f'(2) = 29[/tex]

[tex]f''(2) = 28[/tex]

[tex]f'''(2) = 168[/tex]

The fourth-degree Taylor polynomial is:

P4(x) [tex]= f(2) + f'(2)(x-2) + (f''(2)/2!)(x-2)^2 + (f'''(2)/3!)(x-2)^3 + (f''''(c)/4!)(x-2)x^{2}[/tex]^4

where c is some number between 2 and x.

Using the given third derivative, we can find the fourth derivative:

[tex]f''''(x) = (4x + 1) ^6 * 4[/tex]

Plugging in x = c, we have:[tex]f''''(c) = (4c + 1) ^6 * 4[/tex]

Therefore, the coefficient of [tex](x-2)^4[/tex] in the fourth-degree Taylor polynomial is:[tex](f''''(c)/4!) = [(4c + 1) ^6 * 4] / 24[/tex]

We need to evaluate this at c = 2:[tex][(4c + 1) ^6 * 4] / 24 = [(4*2 + 1) ^6 * 4] / 24 = 18[/tex]

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The system of linear functions for the graph below is given as follows:

The slope-intercept representation of a linear function is given by the equation presented as follows:

y = mx + b

The coefficients of the function and their meaning are described as follows:

For the first line, we have that:

Hence the equation is:

y = -0.5x + 1.

For the second line, we have that:

Hence the equation is:

y = -4x - 6.

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We would need to collect at least 268 service times to return a 95% confidence interval whose width is at most 20 seconds.

(a) We can use the formula for a confidence interval for the mean of a normal distribution with known standard deviation:

CI = X ± z*(σ/√n)

where X is the sample mean, σ is the population standard deviation (in this case, the sample standard deviation is used as an estimate of the population standard deviation since it is known), n is the sample size, and z is the critical value from the standard normal distribution for the desired level of confidence.

For a 95% confidence interval, the critical value is z = 1.96. Plugging in the values, we get:

CI = 8 ± 1.96*(7/√100) = 8 ± 1.372

Therefore, a 95% confidence interval for the mean service time is (6.63, 9.37) minutes.

(b) To find the sample size required to return a 95% confidence interval whose width is at most 20 seconds, we can use the formula for the margin of error:

ME = z*(σ/√n)

where ME is the maximum allowed margin of error (which is 1/3 minute or 0.33 minutes in this case).

Solving for n, we get:

n = (z*σ/ME)^2

For a 95% confidence interval, the critical value is z = 1.96. Plugging in the values, we get:

n = (1.96*7/0.33)^2 ≈ 267.17

Therefore, we would need to collect at least 268 service times to return a 95% confidence interval whose width is at most 20 seconds.

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Answer:

B

Step-by-step explanation:

I put the equations into math-way and it solved the system of equations. X=10 and Y=14.

10 three-point questions and 14 five-point questions

a) There are 3ⁿ ternary digit strings with exactly n digits.

b) There is only 1 string with n digits and n 2's.

c) There are n ternary digit strings with n digits and n-1 2's.



a) For each digit in a ternary digit string, there are 3 possible values (0, 1, or 2). With n digits, you have 3 choices for each digit, giving 3ⁿ total possible strings.

b) If a string has n digits and all are 2's, there's only one possible string, which is '222...2' (with n 2's).

c) If a string has n digits and n-1 of them are 2's, there's one remaining digit that can be 0 or 1. There are n positions this non-2 digit can be in, resulting in n possible strings.

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Answer:

Asiah it’s ego the answer is D

Step-by-step explanation:

I got it right

L ms Brenda

Answer: The answer is Sample D

Step-by-step explanation:

Since f(n) is defined using a primitive recursive function (exponentiation) and follows a recursive structure, we can conclude that f(n) is primitive recursive.

To show that the function f(n) is primitive recursive, we need to demonstrate that it can be defined using basic primitive recursive functions (zero, successor, and projection functions) and can be composed or recursed using only primitive recursive function schemes.

Given the definition of f(n), we can write it as:
- f(0) = 0
- f(1) = 1
- f(2) = 2²
- f(3) = (3³)³
- ...

We can observe that f(n) is defined as a stack of exponentiation operations with the base and the exponent both being n. We can use the following recursive formula to define f(n):

- f(0) = 0


We know that exponentiation is primitive recursive, as it can be defined using multiplication, which is also primitive recursive. We can define exponentiation recursively as:

- exp(a, 0) = 1
- exp(a, b) = a * exp(a, b-1) for b > 0

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Answer:

36.9 degrees

Step-by-step explanation:

Cos(x) = adjacent/hypoteneuse

cos(x) = 12/15

cos(x) = 4/5

x = cos^-1(4/5)

= 36.869898 degrees

= 36.9 degrees

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The percentage change in the interest rate is -80%. This means that the interest rate needs to be reduced by 80% to achieve an increase in the amount of money by 25% in t/2 years time.

The interest on a loan or principal sum can be easily calculated using simple interest. Simple interest is a notion that is employed across a wide range of industries, including banking, finance, automobiles, and more.

Let the original sum of money be P.

According to the question, if P is deposited at a simple interest rate of r% p.a. for t years, it will be doubled. This means that the interest earned on P after t years is P, i.e.,

I = P

The formula for simple interest is I = (P * r * t) / 100. Substituting I = P, we get:

P = (P * r * t) / 100

Simplifying, we get:

r = 100 / t

Now, the question states that if the same amount of money (P) is increased by 25% in t/2 years time, the new amount becomes:

P' = P + (0.25P) = 1.25P

Let the new rate of interest be r'.

The formula for simple interest is I' = (P' * r' * t/2) / 100. Substituting P' = 1.25P, we get:

I' = (1.25P * r' * t/2) / 100

The interest earned on P' after t/2 years is 1.25P - P = 0.25P. Therefore, we have:

I' = 0.25P

Substituting the value of I' in the above equation, we get:

0.25P = (1.25P * r' * t/2) / 100

Simplifying, we get:

r' = 20 / t

The percentage change in the interest rate is given by:

((r' - r) / r) * 100%

Substituting the values of r and r', we get:

((20/t - 100/t) / (100/t)) * 100%

= -80%

Therefore, the percentage change in the interest rate is -80%. This means that the interest rate needs to be reduced by 80% to achieve an increase in the amount of money by 25% in t/2 years time.

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The line passes through the point (4, 5), has its equation of the line to be y = 5/4x

From the question, we have the following parameters that can be used in our computation:

Point, (x, y) = (4, 5)

The equation of a straight line is represented as

y = mx + c

Assuming c = 0,

So, we have

y = mx

This means taht

5 = 4m

So, we have

m = 5/4

Recall that

y = mx

So, we have

y = 5/4x

Hence, the equation of the line is y = 5/4x

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The element at the memory location that is 1 integer behind the memory location pointed to by p2.

(a) *(a1+4) - This expression evaluates to 5. It is equivalent to a1[4].

(b) a1[3] - This expression evaluates to 6, which is the value of the element at index 3 in the array a1.

(c) *p1 - This expression evaluates to 6, which is the value of the element at the memory location pointed to by p1.

(d) *(p1+5) - This expression evaluates to 1, which is the value of the element at the memory location that is 5 integers ahead of the memory location pointed to by p1.

(e) p1[-2] - This expression evaluates to 7, which is the value of the element at the memory location that is 2 integers behind the memory location pointed to by p1.

(f) *(a1+2) - This expression evaluates to 7, which is the value of the element at index 2 in the array a1.

(g) a1[6] - This expression evaluates to 3, which is the value of the element at index 6 in the array a1.

(h) *p2 - This expression evaluates to 7, which is the value of the element at the memory location pointed to by p2.

(i) *(p2+3) - This expression evaluates to 5, which is the value of the element at the memory location that is 3 integers ahead of the memory location pointed to by p2.

(j) p2[-1] - This expression evaluates to 8, which is the value of the element at the memory location that is 1 integer behind the memory location pointed to by p2.

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