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3 years ago

Mrs. Miller wants to estimate the widths the steps in front of her house . Best bench mark for her to use is ?

1 answer:

7 0

The best bench mark to use is an inch. She should use an inch because centimeters are too small and the steps might be less than a foot.

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2x + y - z = -8

Andreyy89

Answer:

Multiply row 1 by $\frac{1}{2}$ .

Step-by-step explanation:

The augmented matrix of the system of linear equation is described below:

$\left[\begin{array}{cccc}2&1&-1&-8\\0&2&3&-6\\-\frac{1}{2} &1&1&-4\end{array}\right]$

Where $a_{11} = 2$ , if we need to create $a_{11} = 1$ , we need to multiply row 1 by $\frac{1}{2}$ , that is to say:

$\left[\begin{array}{cccc}1&\frac{1}{2} &-\frac{1}{2} &-4\\0&2&3&-6\\-\frac{1}{2} &1&1&-4\end{array}\right]$

Hence, the correct answer is: Multiply row 1 by $\frac{1}{2}$ .

5 0

2 years ago

Factor the expression 3w - 6

Musya8 [376]

3 is common and therefore the factorization of the expression will be

3(w-2)

8 0

3 years ago

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using 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.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

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

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

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

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

$1.75 = \frac{X - 3.6}{0.4}$