Simplify.<br><br> 9(y + 7)<br><br> 7 y + 9<br> 63 y<br> 9 y + 63<br> 16 y

This is confusing and makes no sense

Digiron [165]

Answer:

A is the answer

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5 0

3 years ago

Please Help!!!

vaieri [72.5K]

I DON'T UNDERSTAND MATH ANYMORE!!!!!!!!

7 0

3 years ago

What value of n solves the equation?

vampirchik [111]

2^n = 1/8

Try each one of the options in turn

The question marks are negatives right?

7 0

3 years ago

If the level of significance of a hypothesis test is raised from 0.005 to 0.2, the probability of a type ii error will:_________

LiRa [457]

The probability of type II error will decrease if the level of significance of a hypothesis test is raised from 0.005 to 0.2.

<h3 /><h3>What is a type II error?</h3>

A type II error occurs when a false null hypothesis is not rejected or a true alternative hypothesis is mistakenly rejected.

It is denoted by 'β'. The power of the hypothesis is given by '1 - β'.

<h3>How the type II error is related to the significance level?</h3>

The relation between type II error and the significance level(α):

The higher values of significance level make it easier to reject the null hypothesis. So, the probability of type II error decreases.
The lower values of significance level make it fail to reject a false null hypothesis. So, the probability of type II error increases.
Thus, if the significance level increases, the type II error decreases and vice-versa.

From this, it is known that when the significance level of the given hypothesis test is raised from 0.005 to 0.2, the probability of type II error will decrease.

Learn more about type II error of a hypothesis test here:

brainly.com/question/15221256

#SPJ4

7 0

1 year ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago

Simplify. 9(y + 7) 7 y + 9 63 y 9 y + 63 16 y