Is 1 and 3/9 in simplest form?

Which graph represents the function f(x) = 4|x|?

zalisa [80]

Answer:

pick a graph from google

Step-by-step explanation:

there is prablly some u can pick from

4 0

3 years ago

Read 2 more answers

4^2-6(2^x)-16=0 solve for x

ahrayia [7]

Assuming that 4^2-6(2^x)-16=0 is correct, we can rearrange it as:

-6(2^x) + 4^2 - 16=0

Are you sure it's not 6(2^x) + 4^2 - 16=0 ?

If 6(2^x) + 4^2 - 16=0 is correct, then

6(2^x) + 4^2 - 16=16 - 16 = 0, that is, 6(2^x) = 0. Then x = 0 (answer)

8 0

3 years ago

Make r the subject of the formula in 1 - E Rr

Artemon [7]

Answer:

r= 1r- eR = r

Step-by-step explanation:

r-r = ER final answer to equal r= -ER

8 0

3 years ago

Rain is falling steadily in Seattle, Washington. After 6 hours, 4 inches of rain has fallen. How long will it take for 1 inch of

seraphim [82]

You do 6/4, so the answer should be: 1.5 hours!

5 0

3 years ago

in a program designed to help patients stop smoking 232 patients were given sustained care and 84.9% of them were no longer smok

grandymaker [24]

Answer:

$z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869$

$p_v =2*P(Z>1.869)=0.0616$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .

Step-by-step explanation:

1) Data given and notation

n=232 represent the random sample taken

X represent the adults were no longer smoking after one month

$\hat p=0.849$ estimated proportion of adults were no longer smoking after one month

$p_o=0.80$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.8.:

Null hypothesis: $p=0.8$

Alternative hypothesis: $p \neq 0.8$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(Z>1.869)=0.0616$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .

6 0

3 years ago