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3 years ago

Ashley adds Polynomial A and B together:

Mathematics

2 answers:

Juli2301 [7.4K]3 years ago

6 0

Answer:

coefficient for x : 2

coefficient for y : 3

coefficient for z : -7

Step-by-step explanation:

First of all, we have to organize the polynomials and then sum similar terms:

4x^2 - 6y + 5z + 2

4x^2 + 2x +3y - 7z + 0

Hence by focus on the coefficients that multiplies x, y an z we obtain:

coefficient for x : 2

coefficient for y : 3

coefficient for z : -7

hope this helps!!

icang [17]3 years ago

3 0

Answer:

coefficient of x: 2

coefficient of y: 3

coefficient of z: -7

Step-by-step explanation:

To solve this problem, first we need to sum the polynomials A and B, then we need to check the coefficients of x, y and z.

The sum of the polynomials is:

A + B = 5z + 4x^2 - 6y + 2 + 2x + 9y - 12z - 2

A + B = 4x^2 + 2x + 3y - 7z

So, the coefficients are:

coefficient of x^2: 4

coefficient of x: 2

coefficient of y: 3

coefficient of z: -7

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9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operat

Marysya12 [62]

Answer:

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. 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:

$n = 2000, p = 1-0.995 = 0.005$

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

$P(X < 5) + P(X \geq 5) = 1$

We want $P(X \geq 5)$

$P(X \geq 5) = 1 - P(X < 5)$

In which

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

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

$P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044$

$P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445$

$P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235$

$P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480$

$P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765$

$P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710$

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

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3 years ago

After Luca took his dogs for a walk, he gave them 6 dog treats. When Luca's dad got home from work, he gave the dogs t more trea

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Answer:

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Step-by-step explanation:

Equation:

6 + x = 10

Step 1:

x = 10 - 6

Answer:

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Hope This Helps :)

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