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3 years ago

Factor completely 7x3y +14x2y3 − 7x2y2.

Mathematics

2 answers:

netineya [11]3 years ago

8 0

The answer to the question is b

olganol [36]3 years ago

8 0

Answer:

$\largeboxed{7x^2y(x+2xy^2-y)}$

Step-by-step explanation:

$7x^3y+14x^2y^3-7x^2y^2\\\\7x^3y=\boxed7\cdot\boxed{x}\cdot\boxed{x}\cdot{x}\cdot \boxed{y}\\\\14x^2y^3=2\cdot\boxed7\cdot\boxed{x}\cdot\boxed{x}\cdot x\cdot\boxed{y}\cdot y\cdot y\\\\7x^2y^2=\boxed7\cdot\boxed{x}\cdot\boxed{x}\cdot\boxed{y}\cdot y\\\\7x^3y+14x^2y^3-7x^2y^2=\boxed7\cdot\boxed{x}\cdot\boxed{x}\cdot\boxed{y}(x+2\cdot x\cdot y\cdot y-y)\\\\=7x^2y(x+2xy^2-y)$

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The length of a rectangle is 7 ft longer than twice the width. If the perimeter is 92 ft find the length and width of the rectan

user100 [1]

Width of the rectangle is 13 ft and length of the rectangle is 33 ft

<u>Step-by-step explanation:</u>

Step 1:

Given perimeter of the rectangle = 92 ft. Let the width be x, then the length of the rectangle = 2x + 7. Use the formula for perimeter of a rectangle = 2(length + width) to find the dimensions.

92 = 2(2x + 7 + x) = 2(3x + 7) = 6x + 14

⇒ 92 = 6x + 14

⇒ 78 = 6x

⇒ x = 78/6

∴ x = 13 ft

⇒ Width of the rectangle = 13 ft

∴ Length of the rectangle = 2x + 7 = 26 + 7 = 33 ft

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Answer:

Step-by-step explanation:

The data is skewed, and the lowest number of crackers in a package was 7

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3 years ago

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3 years ago

Read 2 more answers

A recent study suggested that 70% of all eligible voters will vote in the next presidential election. Suppose 20 eligible voters

natita [175]

Answer:

0.0479 = 4.79% probability that fewer than 11 of them will vote

Step-by-step explanation:

For each voter, there are only two possible outcomes. Either they will vote, or they will not. The probability of a voter voting is independent of any other voter, which means that the binomial probability distribution is used 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.

70% of all eligible voters will vote in the next presidential election.

This means that $p = 0.7$

20 eligible voters were randomly selected from the population of all eligible voters.

This means that $n = 20$

What is the probability that fewer than 11 of them will vote?

This is:

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)$

In which

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

$P(X = 10) = C_{20,10}.(0.7)^{10}.(0.3)^{10} = 0.0308$

$P(X = 9) = C_{20,9}.(0.7)^{9}.(0.3)^{11} = 0.0120$

$P(X = 8) = C_{20,8}.(0.7)^{8}.(0.3)^{12} = 0.0039$

$P(X = 7) = C_{20,7}.(0.7)^{7}.(0.3)^{13} = 0.0010$

$P(X = 6) = C_{20,10}.(0.7)^{6}.(0.3)^{12} = 0.0002$

$P(X = 5) = C_{20,5}.(0.7)^{5}.(0.3)^{15} \approx 0$

The probability of 5 or less voting is very close to 0, so they will not affect the outcome. Then

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 = 0.0479$

0.0479 = 4.79% probability that fewer than 11 of them will vote

8 0

3 years ago