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3 years ago

14

Solve for x , y , right answers plss !!!!

1 answer:

Tom [10]3 years ago

7 0

Answer:

3 5 6 2 8

Step-by-step explanation:

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Please help with these 3

Margaret [11]

I only know number 9
9. 9x+38

7 0

4 years ago

An English professor assigns letter grades on a test according to the following scheme. A: Top 14% of scores B: Scores below the

NeX [460]

Answer:

Grades between 62 and 64 result in a D grade.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 71.9, \sigma = 7.8$

Find the numerical limits for a D grade.

D: Scores below the top 84% and above the bottom 10%

So below the 100-84 = 16th percentile and above the 10th percentile.

16th percentile:

This is the value of X when Z has a pvalue of 0.16. So X when Z = -0.995.

$Z = \frac{X - \mu}{\sigma}$

$-0.995 = \frac{X - 71.9}{7.8}$

$X - 71.9 = -0.995*7.8$

$X = 64$

10th percentile:

This is the value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 71.9}{7.8}$

$X - 71.9 = -1.28*7.8$

$X = 62$

Grades between 62 and 64 result in a D grade.

6 0

3 years ago

Triangle ACB is the pre-image and triangle DFE is the image. What is the scale factor of the dilation?

Reika [66]

Answer:

½

Step-by-step explanation:

the scale factor of dilation = 6/12 = ½

7 0

3 years ago

Select ALL the correct answers.

Marina CMI [18]

Answer:

The median of A is the same as the median of B.

The interquartile range of B is greater than the interquartile range of A.

Step-by-step explanation:

Given that:

A = number of runs allowed in first 9 games

A = {1, 4, 2, 2, 3, 1, 1, 2, 1}

Rearranging A : 1, 1, 1, 1, 2, 2, 2, 3, 4

Median A = 1/2(n + 1) th term

Median A = 1/2(10) = 5th term = 2

Q1 of A = 1/4(10) = 2.5th term = (1 + 1)/ 2 = 1

Q3 of A = 3/4(10) = 7.5th term = (2+3)/2 = 2.5

Interquartile range = Q3 - Q1 = 2.5 - 1 = 1.5

Number of runs allowed in 10th game = 9

B = {1, 4, 2, 2, 3, 1, 1, 2, 1, 9}

Rearranging B = 1, 1, 1, 1, 2, 2, 2, 3, 4, 9

Median A = 1/2(n + 1) th term

Median A = 1/2(11) = 5.5th term = (2+2)/2 = 2

Q1 of A = 1/4(11) = 2.75th tetm = (1 + 1)/ 2 = 1

Q3 of A = 3/4(11) = 8.25th term = (3+4)/2 = 3.5

Interquartile range = Q3 - Q1 = 3.5 - 1 = 2.5

Median A = 2 ; median B = 2

IQR B = 2.5 ; IQR A = 1.5 ; IQR B > IQR A

7 0

3 years ago

Which expression results from dividing 6x^8-4x^6 by 2x^2

pogonyaev

$\frac{6 {x}^{8} - 4 {x}^{6} }{2 {x}^{2} } = \frac{2 {x}^{6}(3 {x}^{2} - 2) }{2 {x}^{2} } = 2 {x}^{4}(3 {x}^{2} - 2)$

6 0

3 years ago

Read 2 more answers