Mathematics College

Jack has the following scores on exams in his math class 81, 87 93, 89, 61 81 If he wants to make his performance sound as good as possible, which measure should he use the mean, the median, or the mode of his exam grades when he talks about how he is doing? Explain In you response, make sure to a calculate the mean, o calculate the median, and c calculate the mode of his exam grades D compare and pick the measure that makes his performance sound as good as possible. plz explain

Answers

Answer 1

Answer:

mean is 82, Median is 84, mode is 81

Step-by-step explanation:

Answer 2

The mean of the given data set is 82, median is 84 and mode is 81.

The given data set is 81, 87 93, 89, 61, 81.

What is the median?

The median is the value that's exactly in the middle of a dataset when it is ordered.

Now, the ordered pair is 61, 81, 81, 87, 89, 93.

Median = (81+87)/2

=168/2

=84

Mean = Sum of all the observations/Number of observations

= (61 + 81 + 81 + 87 + 89 + 93)/6

= 492/6

= 82

Mode = The mode is the value that appears most frequently in a data set.

Here, 81 is the most frequently in the given data set.

Therefore, the mean of the given data set is 82, median is 84 and mode is 81.

To learn more about the arithmetic mean visit:

https://brainly.com/question/23706022.

#SPJ5

Related Questions

Evaluate the following expression.
(-5+7 x 1)-9x5

Answers

Answer:

Step-by-step explanation:

The answer is -43 ......

[tex]\frac{d}{dx} \int t^2+1 \ dt[/tex]
There is a 2x on the bottom and x^2 on top of the integral symbol
Please help me my teacher did not teach us this:(

Answers

Answer:

[tex]\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2[/tex]

Step-by-step explanation:

[tex]\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?[/tex]

We can use Part I of the Fundamental Theorem of Calculus:

[tex]\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}[/tex]

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

[tex]\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}[/tex]

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

[tex]\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}[/tex]

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

[tex]\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}[/tex]

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

[tex]\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}[/tex]

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

[tex]\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u][/tex]where u represents any function other than a variable

For the first term, replace [tex]\text{t}[/tex] with [tex]2x[/tex], and apply the chain rule to the function. Do the same for the second term; replace

[tex]\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)[/tex]

Simplify the expression by distributing [tex]2[/tex] and [tex]2x[/tex] inside their respective parentheses.

[tex][-(8x^2 +2)] + (2x^5 + 2x)[/tex] [tex]-8x^2 -2 + 2x^5 + 2x[/tex]

Rearrange the terms to be in order from the highest degree to the lowest degree.

[tex]\displaystyle2x^5-8x^2+2x-2[/tex]

This is the derivative of the given integral, and thus the solution to the problem.

If 3 times the difference of x and y is -30, and 2/3 of x plus 1/2 of y is 19, what are x and y

Answers

Answer:

x = 12 and y = 22

Step-by-step explanation:

It is given that,

3 times the difference of x and y is -30

3(x-y) = -30

3x-3y = -30 ....(1)

2/3 of x plus 1/2 of y is 19.

[tex]\dfrac{2x}{3}+\dfrac{y}{2}=19\\\\\dfrac{4x+3y}{6}=19\\\\4x+3y=114\ ...(2)[/tex]

Add equation (1) and (2).

3x-3y + 4x+3y = -30 +114

7x = 84

x = 12

Put the value of x in equation (1)

3x-3y = -30

3(12)-3y = -30

36+30 = 3y

3y = 66

y = 22

Hence, the value of x and y are 12 and 22.

which graph best models y>2x-4

Answers

Sorry, is there an option for this equation?

Answer:the answer is the first graph

(0,-4) with a slope of 2

Step-by-step explanation:

Evaluate 32 divided by 1.6
show work long division