0.1629 nearest tenth (0.1) look at the 0.01 place.

Please help answer needed asap

Basile [38]

Answer:

A

Step-by-step explanation:

a = 3n-30 solve for 'n'

a + 30 = 3n

n = (a+30)/3

n(a) = (a+30)/3

5 0

2 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

-4(6-b)=4 how to solve this problem?

Kaylis [27]

You distribute -4 to each in the parentheses then you get -24 - 4b = 4 then you add 24 on both sides and get -4b = 28 then you divide each side by -4 and it equals -7

3 0

3 years ago

Read 2 more answers

A cyclist estimates that he will bike 80 miles this week. He actually bikes 75.5 miles. What is the percent error of the cyclist

maw [93]

Answer:

<em>The percent error of the cyclist's estimate is 5.63%</em>

Step-by-step explanation:

<u>Percentages</u>

The cyclist estimates he will bike 80 miles this week, but he really bikes 75.5 miles.

The error of his estimate in miles can be calculated as the difference between his estimate and the real outcome:

Error = 80 miles - 75.5 miles = 4.5 miles

To calculate the error as a percent, we divide that quantity by the original estimate and multiply by 100%:

Error% = 4.5 / 80 * 100 = 5.625%

Rounding to the nearest hundredth:

The percent error of the cyclist's estimate is 5.63%

5 0

3 years ago

Please help me thanks!

lions [1.4K]

5 x 16 x 4 = 320 m^3

3 0

3 years ago