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pickupchik [31]
3 years ago
11

There were 7,500 tickets sold for a concert, 20% of which were general admission. How many general tickets were sold?

Mathematics
1 answer:
konstantin123 [22]3 years ago
5 0

7,500*20%=g

7,500*.20=g

1500=g

1,500 general admission tickets were sold.

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Which statements are true about the polynomial 4x3 – 6x2 + 8x – 12?
olga2289 [7]

1. correct- factor: 4

2. correct- factor: 4

3. Incorrect- The polynomial is not prime, and the factored polynomial is 4(2x-3)

4. Incorrect

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If A = 3x power 2+2y + 2 and B=6x power 2 - 8y + 1 then A+B and A-B​
ollegr [7]

Answer:

see explanation

Step-by-step explanation:

Given A = 3x² + 2y + 2 and B = 6x² - 8y + 1 , then

A + B

= 3x² + 2y + 2 + 6x² - 8y + 1 ← collect like terms

= 9x² - 6y + 3

-------------------------------

A - B

= 3x² + 2y + 2 - (6x² - 8y + 1) ← distribute parenthesis by - 1

= 3x² + 2y + 2 - 6x² + 8y - 1 ← collect like terms

= - 3x² + 10y + 1

5 0
3 years ago
Apple would like to estimate the web browsing battery life of the iPad. The following data represents the battery life, in hours
umka21 [38]

Answer:

12.5 hours

Step-by-step explanation:

The data which represents the battery life, in hours, experienced by six users is given below:

10 14 14 16 13 8

To determine the point estimate for the web browsing battery life of the iPad, we find the mean number of hours.

Mean=\dfrac{10+14+14+16+13+8}{6} \\\\=\dfrac{75}{6} \\\\=12.5

The point estimate for the web browsing battery life of the iPad is: 12.5 hours

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}

X - 3.6 = 0.4*1.75

X = 4.3

They last at least 4.3 minutes

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3 years ago
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loris [4]
The answer is 17 cubes.
I drew lines to make cubes so I could see it better 
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3 years ago
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