$32 shorts: 50% discount

Mandarinka [93]

It will be 69 degrees.

(x-1)+x+43=180
2x+42=180
2x=138
x=69

4 0

3 years ago

Sharon paid a total of $25.52, which included sales tax on all the items. What tax rate did Sharon pay? Round the answer to the

Alex787 [66]

The answer would be 7%

4 0

3 years ago

Which is a net for a cube

sladkih [1.3K]

Answer:

wy m by that??!???? im confused

4 0

3 years ago

There are ten shirts in your closet, four blue, three green, and three red. You randomly select a different shirt each day. You

Law Incorporation [45]

It's a probability problem to find the odds of picking a green or red shirt out of the 10 shirts on Thursday, Friday and Saturday since you have randomly already know you have picked a blue shirt on the other days. Not sure if you have this as a multiple question problem as you didn't list any possible answers (A. 7/20, B. 5/47, C. 2/5, D. 4/125) to the question. A, B, C, D being like 7 chances out of 20, 5 chances out 47, 2 chances out of chances 5 or 4 chances out of 125 (example answers only).

Probability = Number favorable outcomes / total number of outcomes

7 0

4 years ago

Read 2 more answers

A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for

tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of t are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

H₀: There is no difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ = 0.

Hₐ: There is a difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

$\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29$

$s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64$

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

$\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11$

$s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08$

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a t-test for two means.

Compute the test statistic value as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}$

$=\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}$

$=-4.123$

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

$m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}$

$=\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}$

$=26.55\\\approx 27$

Compute the critical value for α = 0.05 as follows:

$t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052$

*Use a t-table for the values.

Thus, the critical values of t are ± 2.052.

3 0

3 years ago