Express 11.037 in words

You are playing a game in which a single die is rolled. If a 2 or 5 comes up you win $36. Otherwise, you lose $36. What is your

algol13

1/3 wins and 2/3 losses
given 2 and 5 = 2/6 = 1/3
1, 3, 4, and 6 = 4/6 = 2/3

7 0

3 years ago

G^2 + 4gh; if g = -3 and h = 5

xxTIMURxx [149]

Answer:

Step-by-step explanation:

We know that g = -3 and h = 5

Plugging in the values for g and h you will get:

$g^{2} + 4gh$

$(-3)^{2} + 4(-3)(5)$

$9 + (-60)$

Final answer is $-51$

3 0

3 years ago

How many solutions does -3(2x + 6) = -6x - 18 have

viktelen [127]

Answer: 0=0 The input is an identity or it is true for all values

-3(2x+6)=-6x-18

(-3*2x)+(-3*6)=-6x-18

-6x+-18=-6x-18

0=0

5 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.

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

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