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2 years ago

14

Jacks living room is 16 feet by 20 feet long. if a scale drawing of his living room uses a scale of 3/4 inches= 6 feet, find the

dimentions of his living room on the drawing.

1 answer:

Juli2301 [7.4K]2 years ago

6 0

Answer:

40 inches

Step-by-step explanation:

16×20=320

320÷6=53.33

53.33×3=159.99

159.99÷4=39.9975

39.9975 rounded would be 40

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Jan has 35 teaspoons of chocolate cocoa mix and 45 teaspoons of French vanilla cocoa mix. She wants to put the same amount of mi

spin [16.1K]

Given that <span>Jan has 35 teaspoons of chocolate cocoa mix and 45 teaspoons of French vanilla cocoa mix and that she wants to put the same amount of mix into each jar.

Given that she only wants one flavor mix in each jar and that she wants to fill as many jars as possible.

This question depicts a HCF (highest common factor) question where the maximum amount of jars of each flavor she can fill represent the multiple of the HCF of 35 and 45.

35 = 5 x 7
45 = 5 x 9

Thus the HCF of 35 and 45 is 5.

Therefore, the number of jars of French vanilla cocoa mix Jan will fill is 9.</span>

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3 years ago

Zoe has 8 cups of whole-wheat flour. If each loaf of her favorite bread requires 3/4 of a cup of whole-wheat flour, how many who

Montano1993 [528]

The answer is 10 whole loaves.
We get this answer by dividing 8 by (3/4).
Which results in decimal points as 10.6.
Therefore meaning only 10 loaves were whole.

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3 years ago

Read 2 more answers

What is 15 times 0.85?

Fofino [41]

Answer:

12.75

Step-by-step explanation:

You multiply 15.00 into 0.85

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3 years ago

HEEEEEEEEEEELPPPPPPPPP PLEASEEEEEEE URGENT 15 POINTS+ MORE

slava [35]

Answer:

y - 70, x - 73 should be the answer

Step-by-step explanation:

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2 years ago

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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 question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

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3 years ago