What is the solution to the equation 9^(x

Jose left the white house and drove toward the recycling plant at an average of 40km/h. Rob left some time later driving in the

tiny-mole [99]

I would think it would be 200km before rob caught up with him?

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B33%7D%7B4-%5Csqrt%7B5%7D%20%7D" id="TexFormula1" title="\frac{33}{4-\sqrt{5} }" alt=

kvv77 [185]

The answer I got was: 12 + 3/5

Sorry I didn’t have that symbol thing on my phone so i had to use a slash.
Hope this helps :)

7 0

2 years ago

A Math professor scores all tests with a maximum of 100 points. A student must have an average of 90 on six tests to receive an

nignag [31]

Answer:

No, he cannot

Step-by-step explanation:

Lets say the student takes the maximum socre on his last test, so he gets 100. Now, his average on the six texts will be:

(75+85+90+95+94+100)/6 = 89.83

So, if taking 100 is not enough to get to the average of 90, this means he would need a score greater than 100, which is impossible as 100 is the maximum. So, the student CAN'T get an A

3 0

3 years ago

27. Morgan scored 41 points in 3 games. How many points would you expect him to make in an

kondaur [170]

I think there would be 150 points in 11 games

5 0

3 years ago

Women athletes at the a certain university have a long-term graduation rate of 67%. Over the past several years, a random sample

lana [24]

Answer:

$z=\frac{0.579 -0.67}{\sqrt{\frac{0.67(1-0.67)}{38}}}=-1.193$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.1$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the proportion of women athletes graduated is not significantly lower than 0.67 or 67% at 10% of significance

Step-by-step explanation:

Data given and notation

n=38 represent the random sample taken

X=22 represent the number of women athletes graduated

$\hat p=\frac{22}{38}=0.579$ estimated proportion of women athletes graduated

$p_o=0.67$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is lower than 0.67 or no:

Null hypothesis: $p \geq 0.67$

Alternative hypothesis: $p < 0.67$

When we conduct a proportion test we need to use the z statisitc, 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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.579 -0.67}{\sqrt{\frac{0.67(1-0.67)}{38}}}=-1.193$

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 $\alpha=0.1$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.1$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the proportion of women athletes graduated is not significantly lower than 0.67 or 67% at 10% of significance

3 0

2 years ago

What is the solution to the equation 9^(x - 2) = 27 ?