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kompoz [17]
2 years ago
14

An Internet and cable-television supplier surveyed a random sample of their customers. The results are shown in the table. A 4-c

olumn table with 3 rows. The first column has no label with entries internet, cable television, total. The second column is labeled satisfied with entries 1,820; 824; 2,644. The third column is labeled not satisfied with entries 212, 285, 497. The fourth column is labeled total with entries 2,032; 1,109; 3,141. Which statement about the two-way frequency table is true?
a. The survey represents quantitative data.
b. There is a greater percentage of Internet customers who are not satisfied than cable television customers who are not satisfied.
c. About half of the customers surveyed are cable-television customers.
d. About one-fourth of the cable-television customers are not satisfied.
Mathematics
2 answers:
ale4655 [162]2 years ago
6 0

Answer:

D. About one-fourth of the cable-television customers are not satisfied.

Step-by-step explanation:

Kaylis [27]2 years ago
3 0

Answer:

d. About one-fourth of the cable-television customers are not satisfied.

Step-by-step explanation:

There are 3,141 customers in total, of which 2,032 are internet customers and 1,109 are cable customers.

There are 2,644 satisfied customers, of which 1,820 are internet customers and 824 are cable customers.

There are 497 not satisfied customers, of which 212 are internet customers and 285 are cable customers.

Which statement about the two-way frequency table is true?

a. The survey represents quantitative data.

It is the satisfaction of the customers with the service provided, so it is qualitative data.

b. There is a greater percentage of Internet customers who are not satisfied than cable television customers who are not satisfied.

There are 2,032 internet customers, of which 212 are not satisfied. So 212/2,032 = 10.42% of internet customers are not satisfied.

There are 1,109 cable customers, of which 285 are not satisfied. So 285/1,109 = 25.70% of cable customers are not satisfied.

This statement is not true.

c. About half of the video customers surveyed are cable-television customers.

Of the 3,141 customers surveyed, 1,109 are cable customers. This is 1109/3141 = 35.31%.

This statement is not true.

d. About one-fourth of the cable-television customers are not satisfied.

There are 1,109 cable customers, of which 285 are not satisfied. So 285/1,109 = 25.70% of cable customers are not satisfied.

This is the true statement.

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asambeis [7]

Answer:   c

Step-by-step explanation:

Based on the percent breakdown of 360%.

100% 100% 100% 20%20%20%

15.       15.      15.    3.     3.     3.

100% is all of it, so 15. This is x.

20% of it would be calculated as 0.2 x 15 or 3. This is y.

7 0
2 years ago
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The heights (in cm) and arm spans (in cm) of 31 students were measured. The association between x(height) and y(arm span) is sho
abruzzese [7]

Answer:

27.385 cm longer would we expect Mike's arm span to be than George's.

Step-by-step explanation:

We have to find how many centimeters longer would we expect Mike's arm span to be than George's using the equation:

y= 4.5x + 0.977 x

where y= arm span

and x= height

Given:

Mike's height=x = 175 cm

so Mike's arm span= y= 4.5 x + 0.977 x

                                    = 4.5* (175) + 0.977* (175)

                                    = 787.5 + 170.975

                                    = 958.475 cm

George's height = x = 170 cm

so George's arm span= y= 4.5 x + 0.977 x

                                    = 4.5* (170) + 0.977* (170)

                                    = 765 + 166.09

                                    = 931.09 cm

Mike's arm span longer than George's = Mike's arm span - George's arm span

                                                                 = 958.475 - 931.09

                                                                 = 27.385 cm.

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2 years ago
Evaluate (42)(43)(8-3) 2 .
Dafna1 [17]

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<u>Step 1</u>

1806(8−3)(2)

<u>Step 2</u>

(1806)(5)(2)

<u>Step 3</u>

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6 0
2 years ago
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According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For
SpyIntel [72]

Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

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.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.71

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here n = 16, we want P(X \geq 13). So

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)

In which

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

P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591

P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835

P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273

P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now n = 160. So

\mu = E(X) = np = 160*0.71 = 113.6

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74

Using continuity correction, this is P(X \geq 110 - 0.5) = P(X \geq 109.5), which is 1 subtracted by the pvalue of Z when X = 109.5. So

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

Z = \frac{109.5 - 113.6}{5.74}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

3 0
2 years ago
Paul opened a bakery. The net value of the bakery (in
Umnica [9.8K]

Answer:

1) 2(t - 7)(t + 1)

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x = -b/2a = 6/2 = 3

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