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Sauron [17]
3 years ago
13

Need the answer quick!

Mathematics
1 answer:
elena-s [515]3 years ago
5 0
I think your answer is B. Tell me if its wrong :)
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Just 6-8 pls n thx.......
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6 minus 8 equals -2.
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Given the following table, find the rate of change between f(-1) and f(2)
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I think it’s 7/12 I think it would 7/12
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An insurance company estimates the probability of an earthquake in the next
erma4kov [3.2K]

Answer:

- 27.88

Step-by-step explanation:

Probability of earthquake = 0.0012

P(earthquake). = 0.0012

P(no earthquake) = 1 - p(earthquake) = 1 - 0.0012 = 0.9988

X ____ 60,000 ______ - 100

P(X) ___ 0.0012 _____ 0.9988

The expected value of the policy :

E(X) = Σx*p(x)

E(X) = (0.0012 * 60000) + (0.9988 * - 100)

E(X) = 72 - 99.88

E(X) = - 27.88

3 0
3 years ago
What is the equation of the line that passes through the point (-4,-2) and has a slope of -1/2
vampirchik [111]

Answer: y=-1/2x-4

Step-by-step explanation:

If you use point slope formula you get

y+2=-1/2(x+4)

simplify that in y=mx+b and you get

y=-1/2x-4

4 0
3 years ago
Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list
Radda [10]

Answer:

P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}

Possible values of x: Any from 0 to 5.

P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807

P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015

P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087

P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323

P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this question:

n = 5, p = 0.3, q = 1 - p = 0.7

So

P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807

P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015

P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087

P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323

P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243

8 0
3 years ago
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