Help plz answer needed fast

Mathematics

1 answer:

grigory [225]3 years ago

8 0

We just substitute in -12 for x in each one and see if it's true

2(-12)+6 = -24 + 6 = -18, not 30, NOPE

-0.25(-12) - 1 = 3-1 = 2, that worked, CHECK ME

-12 - 8 = -20, CHECK ME

-35(-12) +2 = 422, not -515, NOPE

3(-12+25)=3(13)=39, CHECK ME

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x varies jointly with the square of y and z. when y=6 and z=5, x=100. Then, what is x when y=8 and z=2?

djyliett [7]

X = k*y^2 * z
100 = 6^2*5*k,
k=0.555
x=0.555zy^2

x=o.555(8^2)(2)=71.1112.

Hope that helps!

8 0

4 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$