All categories

2 years ago

I need c and d answered badly

Mathematics

1 answer:

Kisachek [45]2 years ago

3 0

C. translated down 2 units

---The "-2" located on the outside of f(x) tells us that the y-values are being changed. With that, the graph can be moved up or down. The presence of the negative/subtraction sign tells us that the graph is moved down.

D. translated right 4 units

---The "-4" located with the x in f(x) tells us that the x-values are being changed. The only tricky thing about this is that the direction of the movement is actually the opposite of the sign. So with the negative/subtraction sign, the graph is moved to the right instead of the left.

Hope this helps!

You might be interested in

(4 + 7w2)

ivanzaharov [21]

Answer: 18w^2 - 6w + 4

Explanation:

(4 + 7w^2) - (6w - 11w^2)
= 4 + 7w^2 - 6w + 11w^2
= 18w^2 - 6w + 4

5 0

3 years ago

Which value is an input of the function? –14 –2 0 4

Nina [5.8K]

Answer:

the value is 4.

Step-by-step explanation:

8 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago

Can you give me the answers

Lelu [443]

<h2><em>JUST DO IT YOURSELF ITS BASIC FRACTIONS!!!!!!!!!</em></h2>

7 0

3 years ago

Da + ca<br><img src="https://tex.z-dn.net/?f=da%20%2B%20ca" id="TexFormula1" title="da + ca" alt="da + ca" align="absmiddle" cla

sergejj [24]

Answer:

$\boxed{a(c + d)}$