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4 years ago

X=7w^+4w-6 and y=w^-11w+13

Mathematics

1 answer:

aleksandr82 [10.1K]4 years ago

3 0

Although the terms might seem intimidating, this is a simple substitution problem.

X + Y = _______
Substitute the values of X and Y in terms of w
(7w^2 + 4w - 6) + (w^2 - 11w + 13) = ______
Now all you need to do is combine like terms (terms that have the same variables raised to the same power)
8w^2 - 7w + 7

Final Answer: 8w^2 - 7w + 7

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Enter an expression that represents the fallowing statement and then solve the expression. 2 less than the product of 6 and -10.

IceJOKER [234]

Answer:

-62

Step-by-step explanation:

-2 + (6)(-10) = -2 -60 = -62

2 less than a product is 2 subtracted from the product, or -2 added to the product. A product is formed by multiplying the given numbers.

The product of 6 and -10 is 6×(-10) = -60. 2 less than that is -62.

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3 years ago

Which description can be written as the expression n/4-13

Rom4ik [11]

Order of operations rules are important here: n/4-13

We must do the division first. Thus, n/4-13 translates to "the quotient of n and 4, less 13."

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3 years ago

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Determine the minimum sample size required when you want to be 99% confident that the sample mean is within 0.50 units of the po

vfiekz [6]

Answer:

b. 224

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.58$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

In this problem:

We want to find n, when $M = 0.5, \sigma = 2.9$ So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.5 = 2.58*\frac{2.9}{\sqrt{n}}$

$0.5\sqrt{n} = 7.482$

$\sqrt{n} = 14.964$

$(\sqrt{n})^{2} = (14.964)^{2}$

$n = 223.9$

The nearest integer number to 223.9 is 224.

So the correct answer is:

b. 224

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3 years ago

A consumer group selected 100 different airplanes at random from each of two large airlines. The mean seat width for the 100 air

Novay_Z [31]

Answer:

Decreasing the sample size decreases the width of the confidence interval.

Step-by-step explanation:

-Let s be the sample standard deviation and $\bar X$ be the sample mean.

#We calculate the 95% confidence interval for the sample size 100 as below:

$\bar X\pm z_{0.05/2}\frac{s}{\sqrt{n}}\\\\\\=\bar X\pm 1.960\times\frac{s}{\sqrt{50}}\\\\\\=\bar X\pm 0.1960s$

#we then calculate the confidence interval of sample size =50 and of similar mean and standard deviation:

$\bar X\pm z\frac{s}{\sqrt{n}}\\\\\\=\bar X\pm z_{0.05/2}\times\frac{s}{\sqrt{n}}\\\\\\=\bar X\pm1.960\times \frac{s}{\sqrt{50}}\\\\\\\\=\bar X\pm0.2772s$

#Notice that the smaller sample(n=50) has a narrower confidence interval.

Hence, decreasing the sample size decreases the width of the confidence interval.

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3 years ago

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A clothing design company pays two different people to design its T-shirts.

irina [24]

Answer: D

Step-by-step explanation:

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4 years ago

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