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Sergeeva-Olga [200]

2 years ago

5

The height of a rectangle box is 6 ft. The length is 3 ft longer than the fourfold the width x. The volume is 690 ft cubed. Writ

e an equation in terms of x

1 answer:

Usimov [2.4K]2 years ago

8 0

Answer:

Step-by-step explanation:

18 is the height of it and then you divide 690

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American households spent an average of about $52 in 2007 on Halloween merchandise such as costumes, decorations and candy. To s

Grace [21]

Answer:

The sample mean is$ 52

The claimed population mean is $58

Step-by-step explanation:

American households spent an average of about $52 in 2007 on Halloween merchandise such as costumes, decorations and candy. This tells you that the sample mean is$ 52.

1500 households were surveyed in 2008 and found an average spending of $58 0n Halloween tell us that the claim's population mean is $58

6 0

2 years ago

Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

Ugo [173]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 175$

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that $p = 0.467$ . So

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

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

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

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

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

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

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability, using continuity correction, is $P(X < 12 - 0.5) = P(X < 11.5)$ , which is the pvalue of Z when X = 11.5. So

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

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

Using continuity correction, this probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

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

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0

2 years ago

Solve the following multiplication and division problems. a. 8 T. 1,398 lb. 14 oz. × 6 b. 349 lb. 6 oz. ÷ 130 c. 6 T. 294 lb. ÷

tester [92]

Answer: a) 278382 oz

b) 43 oz

c) 64.1 oz

Step-by-step explanation:

a) 8 T. 1,398 lb. 14 oz. × 6

First we need to change it in 'oz'.

As we know that

$1\ ton=32000\ oz\\\\1\ lb=16\ oz$

so, it becomes,

$8\times 32000+1398\times 16+14\ oz\\\\=278382\ oz$

8 T. 1,398 lb. 14 oz. × 6 becomes

$278382\times 6\\\\=1670292\ oz$

b) 349 lb. 6 oz. ÷ 130

It becomes,

$349\times 16+6\ oz\\\\=5590\ oz\\\\\text{ at last it becomes}\\\\=\frac{5590}{130}\\\\=43\ oz$

c) 6 T. 294 lb. ÷ 3,071

First it becomes,

$6\times 32000+294\times 16\ oz\\\\=196704\ oz$

At last it becomes,

$\frac{196704}{3071}\\\\=64.1\ oz$

Hence, a) 278382 oz

b) 43 oz

c) 64.1 oz

6 0

2 years ago

Read 2 more answers

containers a and b hold 11,875 L of water together. conatainer b holds 2,391L more than container B holds. How many Liters of wa

Licemer1 [7]

Step 1 : 11875=2x+2391 Step 2: 11875-2391=2x Step 3: 9484=2x Step 4: 9484/2=x Step 5: x=4742

4 0

2 years ago

Paul joins a gym that has an initial membership fee and a monthly cost. He pays a total of $295 after three months, and $495 aft

Harrizon [31]

Answer:

a

Step-by-step explanation:

6 0

1 year ago