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Lesechka [4]
3 years ago
11

What are the answers plz help me

Mathematics
1 answer:
Serga [27]3 years ago
8 0

For number 3 it would be |-10| and |10|.

Number 5a is supposed to <, because -1 is greater than -7 when you look at the line graph. The same goes for 5b. 5d is >, because 0 is always greater than -1 and it shows that on the line graph that you have there. 5e is =.

I don't see anything else wrong though. Just the ones I listed.

Hope that helps!

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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.

6 0
3 years ago
PLEASE HELP ME. ONLY HAVE COUPLE MORE MINS LEFT
ludmilkaskok [199]

Answer:

11

Step-by-step explanation:

i dont really know but i think it is 11 because its all two numbers distance

6 0
2 years ago
Hilda has $210 worth of $10 and $12 stock shares. The number of $10 shares is five more than twice the number of $12 shares. How
klasskru [66]
Let's first define variables.
 y: number of shares of $ 10
 x: number of shares of $ 12
 We write the system of equations:
 10x + 12y = 210
 y = 2x + 5
 Solving the system:
 x = 75/17
 y = 235/17
 Answer: 
 she has: 
 75/17 shares of $ 12 
 235/17 shares of $ 10

6 0
3 years ago
Read 2 more answers
Simplify (4a^m+3)^3*(1/16a^1-m)^2
Marina86 [1]

The simplified answer is 27m^2

3 0
2 years ago
3x-1=11 solve for the variable
artcher [175]

First, we should add 1 to both sides to isolate the variable:

3x = 12

Now that x is isolated, we divide by 3:

x = 4

6 0
3 years ago
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