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saveliy_v [14]
3 years ago
11

Write an equation and solve. One telephone company charges $16.95 per month and $0.05 per minute for local calls. Another compan

y charges $22.95 per month and $0.02 per minute for local calls. For what number of minutes of local calls per month is the cost of the plans the same?
A. 16.95 + 0.05m = 22.95 + 0.02m; 200 min

B. 16.95 + 0.05m + 22.95 + 0.02m = 0; 570 min

C. 16.95 + 0.05m = 22.95 + 0.02m; 2 min
Mathematics
2 answers:
frutty [35]3 years ago
6 0
16.95+.05x=22.95+.02x

Subtract .02x from each side and subtract 16.95 from each side

.03x=6
Divide by .03

x=200 minutes.

Your answer is A
jarptica [38.1K]3 years ago
3 0

Answer:

The correct option is A. 16.95 + 0.05m = 22.95 + 0.02m; 200 min

Step-by-step explanation:

Consider the provided information.

One telephone company charges $16.95 per month and $0.05 per minute for local calls.

Let m is represents the number of minutes.

Therefore, the cost of plane can be represent as:

16.95+0.05m

Another company charges $22.95 per month and $0.02 per minute for local calls.

Therefore, the cost of plane can be represent as:

22.95+0.02m

Now equate both the equations:

16.95+0.05m=22.95+0.02m

16.95+0.03m=22.95

0.03m=22.95-16.95

0.03m=6

m=200

For 200 minutes the cost of both plans remains the same.

Hence, the correct option is A. 16.95 + 0.05m = 22.95 + 0.02m; 200 min

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The amount of coffee that a filling machine puts into an 8-ounce jar is normally distributed with a mean of 8.2 ounces and a sta
nordsb [41]

Answer:

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

Step-by-step explanation:

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

Normal probability distribution

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.

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 problem, we have that:

\mu = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018

What is the probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce?

This is the pvalue of Z when X = 8.2 + 0.02 = 8.22 subtracted by the pvalue of Z when X = 8.2 - 0.02 = 8.18. So

X = 8.22

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

By the Central Limit Theorem

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

Z = \frac{8.22 - 8.2}{0.018}

Z = 1.11

Z = 1.11 has a pvalue of 0.8665

X = 8.18

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

Z = \frac{8.18 - 8.2}{0.018}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

0.8665 - 0.1335 = 0.7330

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

8 0
3 years ago
a metal rod 9.4 meters long is cut into two piecew. One piece is 3 times as long as the other m. find the length of the longer p
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x-shorter\ piece\\3x-longer\ piece\\\\x+3x=9.4\\4x=9.4\ \ \ \ \ |:4\\x=2.35\\\\3\cdot2.35=7.05\approx7.1\\\\The\ longer\ piece\ has\ 7.1\ meters
8 0
3 years ago
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