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3 years ago

Timothy evaluated the expression using x = 3 and y = –4. xy-2 3x2y−4

1 answer:

8 0

The expression xy-2, when x=3 and y= -4, becomes (3)(-4)-2, or -14 (answer)

the expresion 3x^2*y - 4 becomes 3(3)^2*(-4) - 4, or 27(-4) - 4, or -108 - 4 = -112 (answer)

Leave Timmy out of this, since it seems he didn't really do anything. ;)

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I NEED HELP REALLY BAD PLEASE ANSWER THIS QUESTION AS SOON AS POSSIBLE!!!! Which of the following expressions represents the dis

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A: 3/4 - 2.

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4 consecutive integers have a sum of 22. What is the least of the 4 integers?

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Answer:

The least of the four integers is 4.

Step-by-step explanation:

Let x be the first integer.

x + (x + 1) + (x +2) + (x + 3) = 22

4x + 6 = 22

4x = 16

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4 + 5 + 6 + 7 = 22

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3 years ago

It is known that 50% of adult workers have a high school diploma. If a random sample of 8 adult workers is selected, what is the

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Answer:

85.56% probability that less than 6 of them have a high school diploma

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have a high school diploma, or they do not. The probability of an adult having a high school diploma is independent of other adults. 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.

50% of adult workers have a high school diploma.

This means that $p = 0.5$

If a random sample of 8 adult workers is selected, what is the probability that less than 6 of them have a high school diploma

This is P(X < 6) when n = 8.

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

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

$P(X = 0) = C_{8,0}.(0.5)^{0}.(0.5)^{8} = 0.0039$

$P(X = 1) = C_{8,1}.(0.5)^{1}.(0.5)^{7} = 0.0313$

$P(X = 2) = C_{8,2}.(0.5)^{2}.(0.5)^{6} = 0.1094$

$P(X = 3) = C_{8,3}.(0.5)^{3}.(0.5)^{5} = 0.2188$

$P(X = 4) = C_{8,4}.(0.5)^{4}.(0.5)^{4} = 0.2734$

$P(X = 5) = C_{8,5}.(0.5)^{5}.(0.5)^{3} = 0.2188$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0039 + 0.0313 + 0.1094 + 0.2188 + 0.2734 + 0.2188 = 0.8556$

85.56% probability that less than 6 of them have a high school diploma

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3 years ago

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4 years ago