A grocery store charges $1.49 per pound of bananas. Use a unit rate to find the price of 3 pounds of bananas.

Robin bought a bunch of packages of gum at the gas station 8 1/7 of a package each week. How Much would she have eaten in 4 week

AlexFokin [52]

32 4/7
Multiply 8 1/7 by4 = 32 4/7

4 0

3 years ago

Read 2 more answers

The diameter and height of this cylinder are equal to the side length, S, of the cube in which the cylinder is inscribed. What i

N76 [4]

Volume of Cylinder = $\pi r^{2}h$
Based on what we're told:
r (radius) = $\frac{s}{2}$
h (height) = $s$
So:
Volume of cylinder = $\pi (\frac{s}{2})^{2}(s)$
= $\frac{1}{2}s^{3} \pi$

6 0

3 years ago

What is 5+5056541652541541343368486479848486463464843658<br> =?

Marta_Voda [28]

5.05654165E45 is your answer

6 0

3 years ago

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

Yuliya22 [10]

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

We can calculate the mean and deviation for eaach group with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =5.92$ represent the sample mean for the group 1

$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

And now we can calculate the statistic:

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

The degrees of freedom are given by:

$df=5+5-2=8$

And now we can calculate the p value using the altenative hypothesis:

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

7 0

3 years ago

Quality control finds on average that 0.026% of the items from a certain factory are detective. One month, 4000 items are checke

telo118 [61]

Given that 4000 items are checked in one month, let the number of detective items be represented as y.

Quality control finds on average that 0.026% of the items in the factory are detective. This implies that

$\text{Number of }\det ective\text{ items = }\frac{\text{0.026}}{100}\times\text{Total number of items checked}$

When 4000 items are checked, we have the number of detective items to be evaluated as

$\begin{gathered} \text{Number of }\det ective\text{ items = }\frac{\text{0.026}}{100}\times\text{Total number of items checked} \\ y=\frac{0.026}{100}\times4000 \\ \Rightarrow y=1.04 \end{gathered}$

Hence, the number of detective items is 1.04.

7 0

1 year ago