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Juli2301 [7.4K]
3 years ago
12

4qu, when q= 5 and u = 7

Mathematics
2 answers:
professor190 [17]3 years ago
4 0

Answer:

140

Step-by-step explanation:

Substitute q = 5 and u = 7 into the expression

4qu

= 4 × 5 × 7 = 20 × 7 = 140

SpyIntel [72]3 years ago
4 0

Answer:

4qu= <em><u>140</u></em>

Step-by-step explanation:

Step 1: Fill in variables                 4(5)(7)

Step 2: Multiply, 4×5                  20(7)

20×7  = 140                          

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Step-by-step explanation:

(4m^2−5)(4m^2+5)

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The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,564 miles, with a standard
DerKrebs [107]

Answer:

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

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 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 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 = 48564, \sigma = 3293, n = 281, s = \frac{3293}{\sqrt{281}} = 196.44

What is the probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct?

This is the pvalue of Z when X = 48101. So

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

By the Central Limit Theorem

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

Z = \frac{48101 - 48564}{196.44}

Z = -2.36

Z = -2.36 has a pvalue of 0.0091

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

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