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2 years ago

Which of the following are not polyhedrons?

Mathematics

1 answer:

Goshia [24]2 years ago

6 0

Answer:C and D

Step-by-step explanation: Hope this helps

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Samara was playing a game. She wrote down this number sentence to find the total number of points for 11 games in a row.

MissTica

Answer:

55 score in his games

Step-by-step explanation:

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3 years ago

How much is a 40% discount off of $74.99

pychu [463]

Answer:

$\frac{40}{100} \times \frac{7499}{100}$

$\frac{2}{5} \times \frac{7499}{100}$

$\frac{1}{5} \times \frac{7499}{50}$

$\frac{7499}{250}$

7499/250= 29.996

74.99-29.996= $44.994

*price after discount

7 0

3 years ago

Stacey is walking laps around a circular track. The radius of the track is 5 ft. What is the total number of feet in one lap?

kobusy [5.1K]

10 pi ft because the formula is 2xradiusxpi

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3 years ago

Help again please smh

mina [271]

Answer:

i think d but i would have thought it would have been divided by 18 but i still think it is d

Step-by-step explanation:

let me know if it helps and if it was right

7 0

4 years ago

Read 2 more answers

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

uysha [10]

Answer:

$(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743$

$(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926$

And we are 9% confidence that the true mean for the difference of the population means is given by:

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$

Step-by-step explanation:

For this problem we have the following data given:

$\bar X_1 = 9$ represent the sample mean for one of the departments

$\bar X_2 = 8$ represent the sample mean for the other department

$n_1 = 25$ represent the sample size for the first group

$n_2 = 20$ represent the sample size for the second group

$s_1 = 2$ represent the deviation for the first group

$s_2 =1$ represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

The confidence given is 95% or 9.5, then the significance level is $\alpha=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df=n_1 +n_2 -2= 20+25-2= 43$

And the critical value for this case is $t_{\alpha/2}=2.02$

And replacing we got:

$(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743$

$(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926$

And we are 9% confidence that the true mean for the difference of the population means is given by:

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$

4 0

3 years ago