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svet-max [94.6K]
2 years ago
10

Given the function {f(x)=32-x}^{2}f(x)=32−x 2 , what is f(-5)f(−5)?

Mathematics
1 answer:
erastova [34]2 years ago
4 0

Step-by-step explanation:

the description is very difficult to understand with special characters and potential typing errors.

I suspect the definition is

function(x) = 32 - x²

function(-5)function(-5) is then very easily calculated.

we need to calculate function(-5) and then square the result, as function(-5)function(-5) is just a simple multiplication of the function results.

function(-5) = 32 - (-5)² = 32 - 25 = 7

so,

function(-5)function(-5) = 7 × 7 = 49

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A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% hi
Mamont248 [21]

Answer:

The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

Step-by-step explanation:

Denote the different kinds of drivers as follows:

L = low-risk drivers

M = moderate-risk drivers

H = high-risk drivers

The information provided is:

P (L) = 0.50

P (M) = 0.30

P (H) = 0.20

Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.

The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:

S = {HHHH, HHHL, HHHM, HHMM}

Compute the probability of the combination {HHHH} as follows:

P (HHHH) = [P (H)]⁴

                = [0.20]⁴

                = 0.0016

Compute the probability of the combination {HHHL} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (L)

               = 4 × (0.20)³ × 0.50

               = 0.016

Compute the probability of the combination {HHHM} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (M)

               = 4 × (0.20)³ × 0.30

               = 0.0096

Compute the probability of the combination {HHMM} as follows:

P (HHMM) = {4\choose 2} × [P (H)]² × [P (M)]²

                 = 6 × (0.20)² × (0.30)²

                 = 0.0216

Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:

P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)

                                                            + P (HHMM)

                                                  = 0.0016 + 0.016 + 0.0096 + 0.0216

                                                  = 0.0488

Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

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Answer:

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Step-by-step explanation:

Since there is an equal number of y's, they cancel out when subtracting. You can put either equation on top. You will get same answer.

<u>To find x:</u>

<u>If you do (3x - 2y = 10) - (5x + 2y = 6​),</u> you will get -2x=4. Divide both sides by -2, and get x = -2.

<u>If you do (5x + 2y = 6​) - (3x - 2y = 10)</u>, you will get 2x=-4.Divide both sides by 2, and get x = -2.

<u>To find y:</u>

Substitute -2 for x in either equation. It doesn't matter.

<u>For 3(-2) - 2y = 10:</u>

3(-2) - 2y = 10 ––Multiply

-6 - 2y =10 ––Add 6 to both sides to get y alone

-2y = 16 ––Divide by -2

y = -8

<u>For 5x + 2y = 6​:</u>

5(-2) + 2y = 6​ ––Multiply

-10 + 2y = 6 ––Add 10 to both sides to get y alone

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y = 8

 

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