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3 years ago

Need help? as soon as possible

Mathematics

1 answer:

Citrus2011 [14]3 years ago

5 0

Answer:

$\sqrt{29}$ =5.385 (must look at the requirement of the test as how many decimal...)

Step-by-step explanation:

the squared is equal to 29 means area=29

square has four side equal.

Area=side * side

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Uhhh help please yup

lina2011 [118]

Answer:

Explanation:

P = 2(W + L)

L = 2x - 3
W = x + 4

P = 2[(x + 4) + (2x - 3)]

So thats the expression for the perimeter

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3 years ago

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Flura [38]

The answer is c. 28.26 cm^2

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3 years ago

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If f (1) = 1 what is f (3)?

Ipatiy [6.2K]

Answer: f (3) = 3

Step-by-step explanation:

If f(1) = 1

Then

f(3) = 3

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3 years ago

Find the midpoint of the segment with the given endpoints.<br> (7,10) and (-1,- 8)

Evgesh-ka [11]

Answer:

(3,1) is the midpoint

Step-by-step explanation:

To find the x coordinate of the midpoint, average the x coordinates of the endpoints

(7+-1)/2 = 6/2 =3

To find the y coordinate of the midpoint, average the y coordinates of the endpoints

(10+-8)/2 = 2/2 = 1

(3,1) is the midpoint

6 0

3 years ago

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Iron deficiency anemia is an important nutritional health problem in the U.S. A dietary assessment was performed on 51 boys 9-11

Bond [772]

Answer:

a) Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

b) $t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

c) $12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163$

$12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837$

Step-by-step explanation:

Information given

$\bar X=12.50$ represent the mean for the daily iron intake

$s=4.75$ represent the sample deviation

$n=51$ sample size

$\mu_o =14.44$ represent the reference value

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

Part a

We want to test if the mean iron intake among the low-income group is different from that of the general population, the system of hypothesis would be:

Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

Part b

Since we don't know the population deviation the statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got

$t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

Part c

The confidence interval would be given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

For the 95% confidence interval we can find the critical value in a t distribution with 50 degrees of freedom and we got:

$t_{\alpha/2}= 2.01$

And replacing we got:

$12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163$

$12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837$

7 0

4 years ago