51 ≤ 15 +6w<br> Solve the inequality for W .<br> Simplify your answer as much as possible.

scoundrel [369]

First of all, find a program or a calculator that will give you the answer. Then you'll know how to make your table. I recommend Desmos. I have inserted the final graph for you.

Make up a table for this graph. Put the lowest point as the central entry in your graph.

x y
-4
-3
-2
-1
0
Now fill in the y values.

x y
-4 1
-3 -3.5
-2 -5
-1 -3.5
0 1

You can use the outline of the parabola and then mark the outline with the points in the table.

8 0

3 years ago

Your school wants to take out an ad in the paper congratulating the basketball team on a successful season, as shown to the righ

Greeley [361]

Answer:

$(7 - x)(4 - x) \\ 28 - 7x - 4x + x {}^{2} \\ x {}^{2} - 11x + 28 = 0 \\ x {}^{2} - 7x - 4x + 28 = 0 \\ x (x - 7) - 4(x - 7 )= 0 \\ (x - 4)(x - 7) = 0 \\$

the value of x is 7 or 4

4 0

3 years ago

Read 2 more answers

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

Helpp me plz respond

spayn [35]

The ratio is 1:6 for this

6 0

3 years ago

The translation of ABCD to A'B'C'D'<br> is given by (×+[?].y-[ ]).

s2008m [1.1K]

Answer:

(x + 1 , y - 3)

Step-by-step explanation:

Compare x-coordinate of A(-6,1) and A'(-5,-2)

-6 + a = -5

a = -5 +6 = 1

x- coordinate of A' = x coordinate of A + 1

Compare y-coordinate of A(-6,1) and A'(-5,-2)

1 + a =-2

a = -2 - 1 = -3

y- coordinate of A' = y coordinate of A - 3

A(-6 , 1) = A'(-6+1 , 1-3) = A'(-5, -2)

D(( -1 , 1) = D'(-1+1, 1 - 3) = D'(0,-2)

C(-2,3) =C'(-2+1, 3-3) = C'(-1,0)

B(-4,3)= B'(-4+1 , 3-3) = B'(-3,0)

7 0

3 years ago

51 ≤ 15 +6w Solve the inequality for W . Simplify your answer as much as possible.