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3 years ago

Use the difference of squares pattern to factor the polynomial.x2 − 16y2 how would I do this

1 answer:

6 0

The general rule for difference of two squares is that when factorised it can be written in the form (a+b)(a-b) where a is the square root of the first part of the function and b is the square root of the second part of the function.
From this we can find the square root of x² to find a and the square root of 16y² to find b.
In this case √x²=x and √16y²=4y
Therefore fully factorised it can be written as (x+4y)(x-4y)

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Solve x^7-11x^6+48x^5-120x^4+225x^3-243x^2-162x+486=0

abruzzese [7]

Answer:

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3 years ago

The local reader's club has a set of 64 hardback books and a set of 24 paperbacks. Each set can be divided equally among the clu

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24/8 = 3, 64/8 = 8
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3 years ago

The mean amount purchased by a typical customer at Churchill's Grocery Store is $27.50 with a standard deviation of $7.00. Assum

Schach [20]

Answer:

a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.

b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00

c) 90% of sample means will occur between $26.1 and $28.9.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 27.50, \sigma = 7, n = 68, s = \frac{7}{\sqrt{68}} = 0.85$

a. What is the likelihood the sample mean is at least $30.00?

This is 1 subtracted by the pvalue of Z when X = 30. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem, we have that:

$Z = \frac{X - \mu}{s}$

$Z = \frac{30 - 27.5}{0.85}$

$Z = 2.94$

$Z = 2.94$ has a pvalue of 0.9984

1 - 0.9984 = 0.0016

0.0016 = 0.16% probability that the sample mean is at least $30.00.

b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?

This is the pvalue of Z when X = 30 subtracted by the pvalue of Z when X = 26.50. So

From a, when X = 30, Z has a pvalue of 0.9984

When X = 26.5

$Z = \frac{X - \mu}{s}$

$Z = \frac{26.5 - 27.5}{0.85}$

$Z = -1.18$

$Z = -1.18$ has a pvalue of 0.1190

0.9984 - 0.1190 = 0.8794

0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00.

c. Within what limits will 90 percent of the sample means occur?

Between the 50 - (90/2) = 5th percentile and the 50 + (90/2) = 95th percentile, that is, Z between -1.645 and Z = 1.645

Lower bound:

$Z = \frac{X - \mu}{s}$

$-1.645 = \frac{X - 27.5}{0.85}$

$X - 27.5 = -1.645*0.85$

$X = 26.1$

Upper Bound:

$Z = \frac{X - \mu}{s}$

$1.645 = \frac{X - 27.5}{0.85}$

$X - 27.5 = 1.645*0.85$

$X = 28.9$

90% of sample means will occur between $26.1 and $28.9.

4 0

3 years ago

Help please with this question

N76 [4]

Answer:

The third graph down

Step-by-step explanation:

-2.1 w > 12.81

Divide each side by -2.1, remembering to flip the inequality

-2.1w / -2.1 < 12.81/ -2.1

w < -6.1

There is an open circle at -6.1 and the line goes to the left

6 0

3 years ago

Read 2 more answers

The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to det

seraphim [82]

Answer:

The required hypothesis to test is $H_0: \mu\leq 30$ and $H_1: \mu>30$ .

Step-by-step explanation:

Consider the provided information.

It is given that the nightclub has recently surveyed a random sample of n = 250 customers of the club.

She would now like to determine whether or not the mean age of her customers is over 30.

Null hypotheses is represents as $H_0$ . Thus the null hypotheses is shown as:

$H_0: \mu\leq 30$

The alternative hypotheses is represents as $H_1$ . Thus the alternative hypotheses is shown as:

$H_1: \mu>30$

Hence, the required hypothesis to test is $H_0: \mu\leq 30$ and $H_1: \mu>30$ .

8 0

3 years ago