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3 years ago

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Mathematics

2 answers:

trasher [3.6K]3 years ago

7 0

Answer:

Step-by-step explanation:

plug the first point in (0), cancels out B, D, E bc g(x) doesnt equal to -1

now the two left are A and C, plug in 1, A doesn't work because 1^2-1=0, not 1

C works because 2(1)-1 = 2-1= 1

Sedaia [141]3 years ago

6 0

Answer:

Step-by-step explanation:

We can see that the y int is -1 which eliminates B D and E.

Next we can test one of the equations by plugging in one of values from table.

From this we can see that C is correct.

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3 years ago

What is (x+y+2)(y+1)<br> a y^2+4xy+2<br> b xy+4y+x+2<br> c y+x+y+2<br> d y^2+xy+x3y+2

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3 years ago

The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

vfiekz [6]

Answer:

a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

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 problem we have that:

$p = 0.75$

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X = 9)$ when $n = 12$ . So

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

$P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581$

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X \geq 16)$ when $n = 20$