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ElenaW [278]
3 years ago
12

A rectangular garage is 15 yards long and 6 yards wide. It costs $8.00 per square yard to put in a new concrete floor. How much

would it cost to put a new concrete floor in the garage?
Mathematics
1 answer:
Lorico [155]3 years ago
6 0

Answer:

multiply 15 by 6 and you will get the area of the garage. multiply that by 8 and you will get the cost

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1/5 of 25 = 25 divided by 5=
vitfil [10]
5 because 25 divided by 5 is 5
5 0
3 years ago
The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assum
Vadim26 [7]

Answer:

a) 0.0951

b) 0.8098

c) Between $24.75 and $27.25.

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 = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762

(a)

What is the likelihood the sample mean is at least $27.00?

This is 1 subtracted by the pvalue of Z when X = 27. So

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

By the Central Limit Theorem

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

Z = \frac{27 - 26}{0.762}

Z = 1.31

Z = 1.31 has a pvalue of 0.9049

1 - 0.9049 = 0.0951

(b)

What is the likelihood the sample mean is greater than $25.00 but less than $27.00?

This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So

X = 27

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

Z = \frac{27 - 26}{0.762}

Z = 1.31

Z = 1.31 has a pvalue of 0.9049

X = 25

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

Z = \frac{25 - 26}{0.762}

Z = -1.31

Z = -1.31 has a pvalue of 0.0951

0.9049 - 0.0951 = 0.8098

c)Within what limits will 90 percent of the sample means occur?

50 - 90/2 = 5

50 + 90/2 = 95

Between the 5th and the 95th percentile.

5th percentile

X when Z has a pvalue of 0.05. So X when Z = -1.645

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

-1.645 = \frac{X - 26}{0.762}

X - 26 = -1.645*0.762

X = 24.75

95th percentile

X when Z has a pvalue of 0.95. So X when Z = 1.645

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

1.645 = \frac{X - 26}{0.762}

X - 26 = 1.645*0.762

X = 27.25

Between $24.75 and $27.25.

3 0
3 years ago
1,743 divided bt 406
Vesna [10]

Answer:

After division we get the value  4.29310344828

Step-by-step explanation:

let x =1,743 ÷406

x= 1,743 ÷406

we get

x=4.29310344828

6 0
3 years ago
Read 2 more answers
-5/3 + 2 1/3 in simplest form
motikmotik

Answer:

4

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
The energy released by metabolism of 1 average candy bar is 1 x 10^6 joules.
Elis [28]

Answer: 32

Step-by-step explanation:

Given: The energy needed for 1 hour of running for an adult =4.5\times10^6\ joules.

Therefore. the energy needed for seven hours of running==7\times4.5\times10^6\ joules.=31.5\times10^6\ joules

Since, he energy released by metabolism of 1 average candy bar =1\times10^6\ joules.

Therefore, the number of candy bars an adult need to eat  to supply the energy needed for seven hours of running.=\frac{31.5\times10^6}{1\times10^6}=31.5\approx32

Hence, the adult need to eat 32 candies to supply the energy needed for seven hours of running.

3 0
3 years ago
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