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tatiyna
3 years ago
13

How many solutions does the equation have? 4(z−5)+2=4z−18 a. 0 b. 1 c. infinitely many

Mathematics
2 answers:
alexandr402 [8]3 years ago
4 0

Answer:

c.Infinitely Many

Step-by-step explanation:

the variable Z could be any number

dolphi86 [110]3 years ago
3 0

Answer:

infinitely

Step-by-step explanation:

Hope i helped (:

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Find the 31st term of the following sequence.. . 9, 15, 21, .... . . 186. 189. 195. . . is there an easy way to do this.
Alja [10]
This problem is an example of an arithmetic series.

The common difference of the second term (15) with the first term (9) and the third term (21) with the second term (15) [15-9=6; 21-15=6] is 6.

The formula for solving this is:
An= A1 +(n-1)*d where An is the nth term, A1 is the first term, n is the number of terms and d is the common difference.

An=9+ (31-1)*6 = 189

The answer is 189.
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4 years ago
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In a given year, the average annual salary of a NFL football player was $189,000 with a standard deviation of $20,500. If a samp
nika2105 [10]

Answer:

15.15% probability that the sample mean will be $192,000 or more.

Step-by-step explanation:

To solve this problem, 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 189000, \sigma = 20500, n = 50, s = \frac{20500}{\sqrt{50}} = 2899.14

The probability that the sample mean will be $192,000 or more is

This is 1 subtracted by the pvalue of z when X = 192000. So

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

By the Central Limit Theorem

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

Z = \frac{192000 - 189000}{2899.14}

Z = 1.03

Z = 1.03 has a pvalue of 0.8485.

1-0.8485 = 0.1515

15.15% probability that the sample mean will be $192,000 or more.

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3 years ago
Samuel earned $18.90 working last weekend. He worked for 24 hours. Which equation and
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