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asambeis [7]
3 years ago
8

Help me solve for 9 and 10 please

Mathematics
2 answers:
Radda [10]3 years ago
8 0
9.
8x-10=x+11
subtract x and add 10 to both sides so that all the x’s are on one side and the numbers on the other
7x=21
Divide both sides by 7 to isolate x
x=3

10.
5x-7=7x-17
Do the same thing as above subtract the 7x and add 7
-2x=-10
divide by -2
x=5
Vika [28.1K]3 years ago
5 0

Answer:

8x-10=x+11

8x-x=11+10

7x=21

x=21/3

x=3

for question 9

5x-7=7x-17

5x-7x=_17+7

-2x=-10

x=-10/-2

x=5

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Answer:

a. The probability is 0.735

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Step-by-step explanation:

If we assume a binomial process, the probability that x customer are willing to switch companies is:

P(x)=nCx*p^{x}*(1-p)^{n-x}

nCx is calculated as:

nCx=\frac{n!}{x!(n-x)!}

Where n are the 20 cable customers and p is the probability 0.2 that the cable customer are willing to switch companies.

Then, P(x) is:

P(x)=20Cx*0.2^{x}*(1-0.2)^{20-x}

The probability that between 2 and 5 (inclusive) customers are willing to switch companies is:

P(2≤x≤5) = P(2) + P(3) + P(4) + P(5)

Where P(2), P(3), P(4) and P(5) are equal to:

P(2)=20C2*0.2^{2}*(1-0.2)^{20-2}=0.1369

P(3)=20C3*0.2^{3}*(1-0.2)^{20-3}=0.2054

P(4)=20C4*0.2^{4}*(1-0.2)^{20-4}=0.2182

P(5)=20C5*0.2^{5}*(1-0.2)^{20-5}=0.1745

So, P(2≤x≤5) is:

P(2≤x≤5) = 0.1369 + 0.2054 + 0.2182 + 0.1745 = 0.735

At the same way, the probability that less than 5 customers are willing to switch is:

P(x<5)=P(0)+P(1)+P(2)+P(3)+P(4)

P(x<5)=0.6296

Finally, the probability that more than 16 customers are willing to switch is:

P(x>16)=P(17)+P(18)+P(19)+P(20)

P(x>16)=0

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