HELP....................

Exit Ticket/CFU/You try:<br> Find the difference: (3 - 2i) - (5 - 6i). Show your work below.

jok3333 [9.3K]

2 Dewey you can still go to get

4 0

3 years ago

Please help with this

Aleksandr-060686 [28]

Answer:

A

Step-by-step explanation:

(3x -14) + 9x is 90 degrees, so if you add 90 it becomes 180 degrees

It is 90 degrees because you can see that there is a right angle there

6 0

3 years ago

Hi, can someone please help me? trying to graduate :(

ehidna [41]

Answer:

D

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

here m = - 2 , then

y = - 2x + c ← is the partial equation

to find c substitute (3, - 2 ) into the partial equation

- 2 = - 6 + c ⇒ c = - 2 + 6 = 4

y = - 2x + 4 ← equation of line in slope- intercept form

7 0

2 years ago

If you know any of these could yoy hep me out :D

Mekhanik [1.2K]

7. X=5
8. X=6
9. B=-3
10. G=3.4

5 0

2 years ago

A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of

Eva8 [605]

Answer:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=250000$ represent the sample mean

$s=37500$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =210250$ represent the value that we want to test

$\alpha=0.1$ 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 to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:

Null hypothesis: $\mu \leq 210250$

Alternative hypothesis: $\mu > 210250$

Compute the test statistic

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".

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

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$

Conclusion

Since is a one right tailed test the p value would be:

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

6 0

3 years ago