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2 years ago

Please solve this with explanations!

Mathematics

1 answer:

lyudmila [28]2 years ago

3 0

Answer:

D. -2x - 8

Step-by-step explanation:

1) Simplify 1/3(6x+15) to 6x+15/3

-6x+15/3 - 3

2) Factor out the common term 3.

-3(2x+5)/3 - 3

3) Cancel 3.

-(2x+5)-3

4) Remove parentheses.

-2x-5-3

5) Collect like terms.

-2x+(-5-3)

6) Simplify

-2x-8

You might be interested in

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

У-3

alisha [4.7K]

Answer:

the answer will be verified by an expert. Until then, talk to a tutor.

5 0

2 years ago

Sketch the graphs of the equations: y = −3 −x

Black_prince [1.1K]

This is a linear equation in slope-intercept form. We can see that the slope is -1 and the y-intercept is -3 -- the line passes through (0, -3).

To graph this, we can simply plot the point (0, -3) and draw a line with slope negative 1 (this means that for every unit right, we have to move 1 unit down.

I have attached a picture of the graph.