Erica went shopping for new clothes for school. She bought a pair of jeans for $35.99 and several shirts for $11.33 each. If x

PLS HELP ME I REALLY NEED HELP I DO NOT FEEL GOOD AND I HAVE TO MCUH WORK...pls? (7th grade work)

Murljashka [212]

Answer:

see below

Step-by-step explanation:

65 - 32 = 13

13 divide by 65 = 0.20

0.20 x 100% = 20%

so the percentage drop is 20%

therefore, Marcus is correct.

5 0

4 years ago

PLS HELP ASAP ILL GIVE BRAINLKEST THANKS

Arisa [49]

Answer:

(4,4)

Step-by-step explanation:

hope this help

4 0

3 years ago

Read 2 more answers

Do bonds reduce the overall risk of an investment portfolio? Let x be a random variable representing annual percent return for t

rjkz [21]

Answer:

COV (all stocks) = 0.55

COV (stocks and bonds) = 0.82

Step-by-step explanation:

Coefficient of Variation is used to measure variability.

It is defined as the ration of standard deviation and the mean.

It can be used to compare variability of two population or two samples.

Formula:

$\text{Coefficient of Variation} = \displaystyle\frac{\text{Standard Deviation}}{\text{Mean}}$

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

x: 14, 0, 39, 25, 32, 27, 28, 14, 14, 15

$Mean = \frac{208}{10} = 20.8$

$Standard~Deviation = \sqrt{\frac{1169.6}{9} } = 11.39$

$Coefficient~of~Variation = \frac{11.39}{20.8} = 0.55$

y = 6, 2, 29, 17, 26, 17, 17, 2, 3, 5

$Mean = \frac{124}{10} = 12.4$

$Standard~Deviation = \sqrt{\frac{924.4}{9} } = 10.13$

$Coefficient~of~Variation = \frac{10.13}{12.4} = 0.82%$

Since coefficient of variation of x is less compared to y, thus it could be said bonds does not reduce overall risk of an investment portfolio.

6 0

3 years ago

How to use an array to find how many rows of 8 are in 40

Lady_Fox [76]

X x x x x x x x
x x x x x x x x
x x x x x x x x
x x x x x x x x
x x x x x x x x

This array shows that there are 5 rows of 8 in 40.

8 0

3 years ago

For a large supermarket chain in a particularâ state, aâ women's group claimed that female employees were passed over for manage

PolarNik [594]

Answer:

1) As the P-value (0.097) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the women's claim that female employees were passed over for management training in favor of their male colleagues.

2) The 95% confidence interval for the population proportion is (0.173, 0.427).

The populations proportion (π=0.4) is included in this confidence interval, so the claim has no enough evidence to be supported.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that female employees were passed over for management training in favor of their male colleagues.

We use the women's part for sample and population proportions.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.4\\\\H_a:\pi$

The significance level is 0.05.

The sample has a size n=50 persons.

The sample proportion is p=0.3.

$p=X/n=15/50=0.3$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.4*0.6}{50}}\\\\\\ \sigma_p=\sqrt{0.0048}=0.069$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.3-0.4+0.5/50}{0.069}=\dfrac{-0.09}{0.069}=-1.299$

This test is a left-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=P(z$

As the P-value (0.097) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that female employees were passed over for management training in favor of their male colleagues.

b) We have to calculate a 95% confidence interval for the proportion.

The sample proportion is p=0.3.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.3*0.7}{50}}\\\\\\ \sigma_p=\sqrt{0.0042}=0.0648$

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.0648=0.127$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.3-0.127=0.173\\\\UL=p+z \cdot \sigma_p = 0.3+0.127=0.427$

The 95% confidence interval for the population proportion is (0.173, 0.427).

The populations proportion (π=0.4) is included in this confidence interval, so the claim has no enough evidence to be supported.

5 0

3 years ago