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garri49 [273]
3 years ago
11

SAT scores are normally distributed, with a mean of 500 points and a standard deviation of 100 points. Suppose you take the SAT.

Several weeks later, you receive your results, which show that you reached the 80th percentile for the math portion.
Recalling that SAT scores are always expressed as multiples of 10, how many points did you get on the test?
Mathematics
2 answers:
Bad White [126]3 years ago
7 0

Answer:

580

Step-by-step explanation:

Apex

OleMash [197]3 years ago
3 0
The solution to the problem is as follows:

 <span>Let be X: scores of SAT 

P(X ≥ x) = 0.9 

P((X - 500)/100 ≥ (x - 500)/100) = 0.9 

P(Z ≥ z) = 0.9 

z = 1.28 

1.28 = (x - 500)/100 

128 + 500 = x 

x = 628
</span>
I hope my answer has come to your help. God bless and have a nice day ahead!
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Divide:(10x²-3x+4)÷(2x-5)
Helen [10]

Answer:

\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}

Quotient: 5x+11

Remainder: 59

Step-by-step explanation:

I'm going to do long division.

The bottom goes on the outside and the top goes in the inside.

  Setup:

        ---------------------------------

2x-5 |  10x^2    -3x      +4

  Starting the problem from the setup:

            5x     +11                        (I put 5x on top because 5x(2x)=10x^2)

        ---------------------------------   (We are going to distribute 5x to the divisor)

2x-5 |  10x^2    -3x      +4

        -(10x^2  -25x)                  (We are now going to subtract to see what's left.)

      -----------------------------------

                      22x      +4          (I know 2x goes into 22x, 11 times.)

                                                ( I have put +11 on top as a result.)

                    -(22x     -55)        (I distribute 11 to the divisor.)

                 -----------------------

                                  59          (We are done since the divisor is higher degree.)

The quotient is 5x+11.

The remainder is 59.

The result of the division is equal to:

5x+11+\frac{59}{2x-5}.

We can actually use synthetic division as well since the denominator is linear.

Let's solve 2x-5=0 to find what to put on the outside of the synthetic division setup:

2x-5=0

Add 5 on both:

2x=5

Divide both sides by 2:

x=5/2

Or realize that 2x-5 is the same as 2(x-(5/2)) which you will have to do anyways if you choose this route:

So 5/2 will go on the outside:

5/2  |    10          -3            4

      |                  25         55

         ------------------------------

           10          22        59

So we have:

\frac{10x^2-3x+4}{2x-5}

=\frac{10x^2-3x+4}{2(x-\frac{5}{2})}=\frac{1}{2} \cdot \frac{10x^2-3x+4}{x-\frac{5}{2}}=\frac{1}{2}(10x+22+\frac{59}{x-\frac{5}{2}})

Distribute the 1/2 back:

\frac{10x^2-3x+4}{2x-5}=\frac{10x+22}{2}+\frac{59}{2(x-\frac{5}{2})}

\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}

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Answer:

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