A spinner has 10 equally sized sections, 3 of which are green and 7 of which are yellow. The spinner is spun and, at the same time, a fair coin is tossed. What is the probability that the spinner lands on green and the coin toss is heads?

The profit for one small cake is £1.31.

Let's first determine the overall count of parts demonstrated in the ratio 5:7:12, which is equal to 24. Subsequently, each part would hence correspond to a quantity of nine cakes as per the equation below:

216 cakes / 24 = 9 cakes

5 parts x 9 cakes/part = 45 cakes

Likewise, similarly calculating the number of medium cakes leads us to 63 cakes, while that of large cakes sums up to 108 cakes:

7 parts x 9 cakes/part = 63 cakes

12 parts x 9 cakes/part = 108 cakes

Considering what the profit for one small cake is denoted by "x", and similarly, the profit for one medium cake is "2x" and the profit for one large cake is "3x", we can express the total gain as follows:

45x + 63(2x) + 108(3x) = 648.45

After similifying and solving for "x", we obtain an answer of £1.31 after rounding it off to two decimal places:

495x = 648.45

x = 1.31 (rounded to two decimal places)

Thus, the income earned from selling one single small cake is estimated at £1.31.

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Answer: 2 Radians

Step-by-step explanation:

    A full circle is 2 full radians, also known as 2π. Half of a circle is equal to 180 degrees, which is equal to one radian, π.

    I have attached a picture of the unit circle below, this picture comes form www.remind.com, but you can find many online.

The limits which describe the end behavior of the function are given as follows:

The end behavior of a function refers to how the function behaves as the input variable (typically denoted as x) approaches positive or negative infinity. In other words, the end behavior describes the long-term behavior of the function as x becomes very large (either positively or negatively).

We can see that the left of the graph approaches y = 1, hence the limit is given as follows:

[tex]\lim_{x \rightarrow -\infty} f(x) = 1[/tex]

The right of the graph approaches y = 0, hence the limit is given as follows:

[tex]\lim_{x \rightarrow \infty} f(x) = 0[/tex]

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The population predicted after two hours is approximately 666.5 bacteria as they have exponential growth.

The continuous exponential growth model is given by the formula:

P(t) = P0 * [tex]e^{(rt)[/tex]

where P0 is the initial population, r is the relative growth rate (expressed as a decimal), t is the time in hours, and P(t) is the predicted population after t hours.

In this case, P0 = 535, r = 0.12 (12% per hour), and t = 2. Plugging in these values, we get:

P(2) = 535 * [tex]e^{(0.12*2)[/tex] ≈ 666.5

Therefore, the population predicted after two hours is approximately 666.5 bacteria (rounded to the nearest tenth).

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Answer: -0.25

Step-by-step explanation:

Start by undoing with PEMDAS (order of operations, P-parenthesis, E-exponents, etc)

+3 on both sides of equal sign, which gives you

8n=-2

then divide by 8 on both sides (undo the times 8)

n=-0.25

Answer: -0.25

Step-by-step explanation:

Start by undoing with PEMDAS (order of operations, P-parenthesis, E-exponents, etc)

+3 on both sides of equal sign, which gives you

8n=-2

then divide by 8 on both sides (undo the times 8)

n=-0.25

Answer:

Step-by-step explanation:

Z:

the angle b/n z and  65 is 90 b/c its a square and they are found on a straight line (180)

- 65 + 90 + z = 180

-155 + z = 180

-z= 180 - 155= 25

x:

-x + 90 = 180

- x= 180 - 90 = 90

y:

in this case now we have an inner triangle which has only one missing angle so as we know for any triangle ABC; <A + <B + <C = 180 so

- y + x + 65=180

-y+ 90 + 65= 180

-y=180 - 155

- y = 25

Answer: 288.62

Step-by-step explanation:

10.4^2 + 15.3^2 = 342.25,

c^2 = 342.25

c (hypotenuse of the right triangle & length of rectangle) = 18.5

7(8) + 10.4(15.3) = 288.62

Tom can choose from 7,000 possible schedules if every possible course is available at the time he's registering.

Tom can choose 22,400 possible schedules when the number of acceptable courses is doubled. The number of possible schedules is increased by a factor of approximately 3.2

(a) If every possible course is available at the time Tom is registering, he can choose from:

5 choices for English x 5 choices for History x 5 choices for Math/Stats x 8 choices for Computer Science x 7 choices for general Science

= 5 x 5 x 5 x 8 x 7

= 7000 possible schedules.

(b) If the number of acceptable courses in each of the five areas is doubled, Tom can choose from:

10 choices for English x 10 choices for History x 10 choices for Math/Stats x 16 choices for Computer Science x 14 choices for general Science

= 10 x 10 x 10 x 16 x 14

= 22400 possible schedules.

The number of possible schedules is increased by a factor of:

22400 / 7000 = 3.2 (approximately)

So the number of possible schedules is increased by a factor of approximately 3.2 when the number of acceptable courses is doubled in each of the five areas.

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See text below

While making up his schedule for spring semester, Tom complains that he doesn't have very many choices of schedule because of the general education requirements he has to meet. His advisor tells Tom that he has to take one course from each of English (5 choices), History (5 choices), Math/Stats (5 choices), Computer Science (8 choices), and general Science (7 choices). Does Tom have a legitimate gripe?

(a) If every possible course is available at the time he's registering, how many possible schedules can he choose from (disregarding when the classes meet)?

Tom can choose   -------- from possible schedules.

(b) In an unprecedented effort to make the general education requirements more accessible, the dean of Tom's college decides to double the number of acceptable courses in each of those five areas. What effect does this have on the number of possible schedules?

Tom can choose ----------- from possible schedules when the number of acceptable courses are doubled. The number of possible schedules is increased by a factor of-------

Answer:

picture

Step-by-step explanation:

The distance between each big tick mark is 1, (for example take 2 and 1, 2 minus 1 equals to 1), however, in between each of these big tick marks are four small tick marks. The distance between each small tick mark is the distance between each big tick mark divided by the amount of small tick marks plus 1 (1 divided by 4, which is 1/4).

So in this number line you count by 1/4.

Answer:

We need to find the percentage of tomato plant that has a height less than or equal to 3.88 feet.

We know that the tomato plant heights are normally distributed with a mean of 3.56 ft and a standard deviation of 0.25 ft.

Thus, the percentage P(x≤3.88) of the heights being less than 3.88 can be calculated using the z-score z, given

Answer

About 89.97% of tomato plants.

Answer:

89.97%

Step-by-step explanation:

Answer:

110.5

Step-by-step explanation:

The first step is to find the measure of angle RVT

Angle Formed by Two Chords= 1/2(SUM of Intercepted Arcs)

RVT = 1/2 ( 97+42)

RVT =1/2 ( 139)

RVT = 69.5

Now RVT and RVU equals a straight line so they add to 180

RVT + RVU = 180

69.5 + RVU = 180

RVU = 110.5

The scale she use for the axes on her scatter plot are:

For the y-axis she should use a scale from 0 to 180 in 10-dollar intervals.

For the x-axis she should use a scale from 0 to 8 in one-hour intervals.

A scatter plot is a type of graph that displays values for two variables as a set of points on a two-dimensional coordinate system.

These are the appropriate scales for the axes on Caroline's scatter plot as they will allow for a clear visualization of the relationship between the number of hours spent in the mall and the amount of money spent.

The y-axis should show the amount of money spent, and a scale from 0 to 180 in 10-dollar intervals will provide enough range to accommodate the highest value in the data set (175 dollars) and allow for clear and easy-to-read axis labels.

Similarly, the x-axis should show the time spent in the mall, and a scale from 0 to 8 in one-hour intervals will also provide enough range to accommodate the highest value in the data set (7 hours) and allow for clear and easy-to-read axis labels.

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Answer:

The answer is A. $1,166.40

Step-by-step explanation:

350 is the answer

Step-by-step explanation:

I hope it helps

The probability that the distance between two points on a circle of radius 1 is at most 1.5 is approximately 0.477 or 47.7%, which corresponds to the range of central angles less than or equal to 3 radians out of the total central angle of 2π radians.

Let's consider two points on a circle of radius 1, labeled A and B. The distance between A and B is given by the length of the arc AB on the circle, which is a fraction of the circumference of the circle. The circumference of a circle of radius 1 is 2π, so the length of the arc AB is given by

d = θ/2π * 2πr = θ

where d is the distance between A and B, θ is the central angle between A and B (in radians), and r is the radius of the circle (which is 1 in this case).

To find the probability that the distance between A and B is at most 1.5, we need to find the range of central angles that correspond to arc lengths less than or equal to 1.5. Let's call this range of angles θ₁. We have

θ₁ = 2d/r = 2(1.5)/1 = 3

This means that the central angle between A and B must be less than or equal to 3 radians for the distance between A and B to be at most 1.5. Since the total central angle of the circle is 2π radians, the probability of selecting two points with a central angle less than or equal to 3 radians is

P(θ ≤ 3) = θ/2π = 3/2π ≈ 0.477

Therefore, the probability that the distance between two points on a circle of radius 1 is at most 1.5 is approximately 0.477 or 47.7%.

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The critical value t* for a two-tailed 95% confidence interval with 19 degrees of freedom is approximately 2.093.

The critical value t* depends on the degrees of freedom (df) and the desired level of confidence. For a 95% confidence interval and a sample size of n=20, the degrees of freedom would be df = n-1 = 19.

Using a t-distribution table with 19 degrees of freedom (since n-1=20-1=19), we can find the value of t* that corresponds to a 95% confidence level. For a two-tailed test (which is typically used for a confidence interval), the area in each tail is 0.025 (since 1 - 0.95 = 0.05, and we need to split this evenly between the two tails).

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Answer:

Volume L× W × H

Volume 4 × 3 × 2

4 × 3 × 2 =24

If the fish poke is 6 feet long and she has a rectangular prism of 24 since 4 × 3 ×2 equals 24 then yes she can store the fishing pole in the box

24

Answer:

y- intercept = - 14

Step-by-step explanation:

the y- intercept is the point on the y- axis where the line crosses.

at any point on the y- axis, the x- coordinate is zero.

let x = 0 in the equation and solve for y , the y- intercept

5(0) - 2y = 28

0 - 2y = 28

- 2y = 28 ( divide both sides by - 2 )

y = - 14 ← y- intercept

Answer:

a. 45°

b. Arc KPR and Arc RLP

c. Arc PR and Arc RL

d. Arc LQ, Arc QR, and Arc RP

Step-by-step explanation:

a. 90/2=45°

b. Observation

c. They both have measures of 90 degrees

d. Arc Addition Postulate

Answer:

see below

Step-by-step explanation:

angle lcr appears to be a right angle

therefore if a line bisects LCR it would split the angle into 2 congruent halves:

so

90/2=45

m<LCR=45

b) arc KP and arc PR

c)

arc KL and arc LQ

d) arc PR, arc RQ and arc QL

If Mrs. Smith's oatmeal cookie recipe calls for 2/5 of a cup of sugar to make a full batch of cookies, then to make half a batch of cookies, she will need half the amount of sugar.

So, to find out how much sugar Mrs. Smith needs to make half a batch of cookies, we can use the following calculation:

Sugar needed for half a batch = (2/5 cup sugar) x (1/2 batch)

Simplifying this expression, we get:

Sugar needed for half a batch = 2/10 cup sugar

Which can be further simplified to:

Sugar needed for half a batch = 1/5 cup sugar

Therefore, Mrs. Smith will need 1/5 cup of sugar to make half a batch of oatmeal cookies.

Perimeter = 28 yards

Area 49 square yards.

Answer:

[tex]2 \dfrac{1}{2}[/tex]     OR     [tex]\dfrac{5}{2}[/tex]

Step-by-step explanation:

We can see that there are 2 whole shaded circles, and there is 1 half shaded circle.

Therefore, we can represent the shaded area as [tex]\bold{2 \dfrac{1}{2}}[/tex] circles.

We can convert this mixed number to an improper fraction by multiplying the whole number by the denominator and adding the resulting value to the numerator.

[tex]2 \dfrac{1}{2}[/tex]

[tex]= \dfrac{(2 \times 2) + 1}{2}[/tex]

[tex]= \dfrac{5}{2}[/tex]

Note that the best name for the line segment indicated is  [tex]\overleftrightarrow {AB}[/tex]

A line segment is a section of a straight line that is bounded by two different end points and contains every point on the line between them. The Euclidean distance between the ends of a line segment determines its length.

Other types of lines are:

Lines that are parallel. Two lines in the same plane, either intersecting or not.

Lines that cross. Intersecting lines are two or more lines that cross in a plane.

Lines that are concurrent. Concurrent lines are defined as numerous lines that cross through each other at the same place.

Points that are curved.

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Full Question:

See attached image.

The domain of the ellipse is [-7, 7].

The range of the ellipse is [-4, 4].

In mathematics, the domain of a function is the set of all possible input values for which the function is defined. In other words, it is the set of all values of the independent variable (usually denoted by x) for which the function produces a valid output (usually denoted by y).

In mathematics, the range of a function is the set of all possible output values that the function can produce for a given set of input values. In other words, it is the set of all values of the dependent variable (usually denoted by y) that can be obtained from the function.

To find the domain and range of an ellipse, we need to identify the minimum and maximum x-coordinates and y-coordinates of the points on the ellipse.

The center of the ellipse is the midpoint of its major axis, which is the line segment connecting its vertices (-7,0) and (7,0). The midpoint is ((-7+7)/2, (0+0)/2), or (0,0). This means the center of the ellipse is at (0,0).

The semi-major axis is the distance from the center of the ellipse to one of its vertices, which is 7. The semi-minor axis is the distance from the center of the ellipse to one of its co-vertices, which is 4.

So, the equation of the ellipse can be written in standard form as:

(x - 0)²/7² + (y - 0)²/4² = 1

Simplifying, we get:

x²/49 + y²/16 = 1

To find the domain and range, we need to consider the x-coordinates and y-coordinates separately.

For the x-coordinates, we can rearrange the equation as:

x² = 49 - 49y²/16

This tells us that the minimum and maximum values of x occur when y is at its extremes, which are -4 and 4. Substituting these values, we get:

x² = 0 and x² = 49

Taking the square root, we get:

x = 0 and x = ±7

So, the domain of the ellipse is [-7, 7].

For the y-coordinates, we can rearrange the equation as:

y² = 16 - 16x²/49

This tells us that the minimum and maximum values of y occur when x is at its extremes, which are -7 and 7. Substituting these values, we get:

y² = 0 and y² = 16

Taking the square root, we get:

y = 0 and y = ±4

So, the range of the ellipse is [-4, 4].

Therefore, the domain of the ellipse is [-7, 7], and the range of the ellipse is [-4, 4].

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Answer:  Option C

Step-by-step explanation: To compare the rates at which the pipes are leaking, we can find the amount of water each pipe leaks per minute.

For Pipe A:

3/4 gallon ÷ 30 minutes = 1/40 gallon per minute

For Pipe B:

1 1/4 gallon ÷ 45 minutes = 5/36 gallon per minute

For Pipe C:

1/2 gallon ÷ 20 minutes = 1/40 gallon per minute

Therefore, Pipes A and C are leaking at the same rate, which means option C is true. Option A and D are false, as Pipe B has the fastest leak rate. Option B is false, as both Pipe A and Pipe C are leaking at the same rate.

The initial investment will be doubled after approximately 14.6 years.

The formula for continuous compounding is given by the formula A = Pe^(rt). We we want to find the time it takes for the initial investment to double, so we can set A = 2P and solve for t:

2P = Pe^(rt)

e^(rt) = 2

Taking the natural logarithm of both sides, we get:

rt = ln(2)

t = ln(2)/r

Substituting the values:

t = ln(2)/0.0475

t = 14.5925722223

t ≈ 14.6 years.

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Answer:

Step-by-step explanation:

A is correct

Answer: it is more probable that Peter will finish ahead of Francis and Christopher than that both Anthony and Peter will receive ribbons.

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Step-by-step explanation: