ASAP help me out plz

72 is the product of Mai's height and 6.<br>

7nadin3 [17]

Answer:

12

Step-by-step explanation:

72/6 = 12

Or 72= h*6

Be sure to mark me brainliest!

3 0

3 years ago

Read 2 more answers

F(x)=4x squared −7x+7, Find f(2)

Lana71 [14]

Answer: 9

Step-by-step explanation:

4x²-7x+7 f(2)=4*(2)^2-7(2)+7=16-14+7=9

7 0

3 years ago

Refer to the data in the logic circuit waferexample on page 298 ( ẋ=0.16, s²=.000314,n=15). Should the industrial engineer accep

muminat

Answer:

a) If we compare the p value and a significance level assumed $\alpha=0.01$ we see that $p_v$ so we can reject the null hypothesis, and the the actual true mean is significantly lower than 0.18 ounces.

b) The rejection region zone should be: t <-2.62

c) The rejection region in terms of the mean is: $\bar x$ <0.168

Step-by-step explanation:

Data given and notation

$\bar X=0.16$ represent sample mean

$s=\sqrt{0.000314}=0.0177$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =0.18$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is less than 0.18 ounces, the system of hypothesis would be:

Null hypothesis: $\mu \geq 0.18$

Alternative hypothesis: $\mu < 0.18$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{0.16-0.18}{\frac{0.0177}{\sqrt{15}}}=-4.37$

Calculate the P-value

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

$p_v =P(Z$

Conclusion

If we compare the p value and a significance level assumed $\alpha=0.01$ we see that $p_v$ so we can reject the null hypothesis, and the the actual true mean is significantly lower than 0.18 ounces.

Should the industrial engineer accept or reject the null hypothesis that μ>= 0.18 ounce?

We reject the null hypothesis.

Rejection region in terms of t: t <

On this case we need to find first the degrees of freedom given by:

$df=n-1=15-1=14$

Now since the test it's one left tailed test we need a value on the t distribution with 14 degrees of freedom such that we have 0.01 of the area on the left and 0.99 of the area on the right. For this case we can use the following excel code:

"=T.INV(0.01,14)" and we got that the rejection region zone should be: t <-2.62

Rejection region in terms of ẋ: ẋ < .

We can use th critical value founded on the last part and we can use this formula similar to the z score.

$t=\frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

And we can solve for $\bar X$ like this:

$\bar X =\mu + t*\frac{s}{\sqrt{n}} =0.18-2.62*\frac{0.0177}{\sqrt{15}}=0.168$

And the rejection region in terms of the mean is: $\bar x$ <0.168

5 0

3 years ago

What are the transformations for this quadratic function y= 3 (x+4) power of 2 -8

lianna [129]

Answer:

you can use math solver to scan

Step-by-step explanation:

just try

6 0

3 years ago

There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit

DaniilM [7]

Answer: p = 0.9337

Step-by-step explanation: from the question, we have that

total number of pen (n)= 15

number of pen that has never been used=9

number of pen that has been used = 15 - 9 =6

number of pen choosing on monday = 3

total number of pen choosing on tuesday=3

note that the total number of pen is constant (15) since he returned the pen back .

probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

probability of picking a pen that has been used on tuesday = 6/15 = 2/5

probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5

on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .

we know that

probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.

to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution

p(x=r) = $nCr * p^{r} * q^{n-r}$

n= total number of pens=15

r = number of pen chosen on tuesday = 3

p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

by slotting in the parameters, we have that

p(x=3) = $15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}$

p(x=3) = 455 * $0.4^{3} * 0.6^{12}$

p(x=3) = 455 * 0.064 * 0.002176

p(x=3) = 0.0633

thus probability that 3 of the pens has been used on tuesday. = 0.0633

probability of ball that none of the 3 pen has never being used on tuesday = 1 - 0.0633 = 0.9337

3 0

3 years ago