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Dovator [93]
3 years ago
7

HELPPPPP MEEEEEE PLSSSSS ASAPPPPPPPPP!!!!!!!!

Mathematics
2 answers:
mestny [16]3 years ago
7 0

Answer:

6. (Second page)

A. 1/4 of 24 = 6

B. 2/3 x 18 = 12

C. Cannot read problem

D. 3/4 x 25 = 18.75

E. 1/5 x 45 = 9

F. 4/5 x 45 = 36

Sever21 [200]3 years ago
5 0

Answer:

2. B

3.B

4.C

6A. 6

6B. 12

6C. 19.6

6D. 18.75

6E. I cant see

6F. I cant see

Step-by-step explanation:

2. 1/4*40=10

3. type the choice into the calculator

4. 2/5 of 30 or 2/5*30=12

6. A-F type into the calculator

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The average breaking strength of a certain brand of steel cable is 2000 pounds, with a standard deviation of 100 pounds. A sampl
Westkost [7]

Answer:

1963.2 pounds (lbs.)

Step-by-step explanation:

Things to understand before solving:

  • - <u>Normal Probability Distribution</u>
  1. The z-score formula can be used to solve normal distribution problems. In a set with mean ц and standard deviation б, the z-score of a measure X is given by: Z=\frac{X-u}{a}

The Z-score reflects how far the measure deviates from the mean. After determining the Z-score, we examine the z-score table to determine the p-value associated with this z-score. This p-value represents the likelihood that the measure's value is less than X, or the percentile of X. Subtracting 1 from the p-value yields the likelihood that the measure's value is larger than X.

  • - <u>Central Limit Theorem</u>
  1. The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean ц and standard deviation б , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean ц  and standard deviation s=\frac{a}{\sqrt{n} }

As long as n is more than 30, the Central Limit Theorem may be applied to a skewed variable. A specific kind of steel cable has an average breaking strength of 2000 pounds, with a standard variation of 100 pounds.

This means, ц  = 2000 and б = 100.

A random sample of 20 cables is chosen and tested.

This means that n = 20, s=\frac{100}{\sqrt{120} } =22.361

Determine the sample mean that will exclude the top 95 percent of all size 20 samples drawn from the population.

This is the 100-95th percentile, or X when Z has a p-value of 0.05, or X when Z = -1.645. So Z=\frac{X-u}{a}

  • By the Central Limit Theorem

Z=\frac{X-u}{a} \\-1.645=\frac{X-2000}{22.361} \\X-2000=-1.645*22.361

X =1963.2

<h3>Answer:</h3>

The sample mean that will cut off the top 95% of all size 20 samples obtained from the population is 1963.2 pounds.

3 0
3 years ago
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