Using ratio pleases help i´m stuck i need #5 for one side 1-5 for the other

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dalvyx [7]

Answer:

Step-by-step explanation:

Mathematicians stand on each other's shoulders. — Carl Friedrich Gauss We must admit with humility that, while number is purely a product of our minds, space has a reality outside our minds, so that we cannot completely prescribe its properties a priori.

4 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

Can someone help me pls :)

Makovka662 [10]

Answer:a

Step-by-step explanation:

Can I have brainlest

3 0

3 years ago

Read 2 more answers

Emma kept track of the number of puzzles solved during the past 10 days. She realized that she didn't solve any puzzles in the l

AveGali [126]

I think the answer is 13

6 0

3 years ago

A company uses the declining-balance method of calculating depreciation expense. On January 1, the company buys machinery for $7

NeX [460]

Answer:

$516,000

Step-by-step explanation:

Given:

- Beginning boo value of asset A = $750,000

- The salvage value of the asset B = $100,000

- The estimated life of asset t = 10 years

Find:

What is the book value in year 3?

Solution:

- Under the declining balance method, double declining method depreciation rate (R) is calculated by:

R = 1 / t

R = 1 / 10 = 10%

- The accumulated depreciation for 3 years is as follows:

Depreciation: ( A - B ) * R * 2

Year 1: DDB = ( 750,000 - 100,000 ) * 0.2 = $130,000

Year 2: DDB = ( 620,000 - 100,000)*0.2 = $104,000

- The book value in year 3 is:

$ 620,000 - $104,200 = $516,000

7 0

3 years ago