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MAXImum [283]
2 years ago
7

Which expression is equivalent to 1.5 raised to the fifth power divided by 0.7 raised to the fourth power, all raised to the thi

rd power?
1.5 raised to the fifteenth power divided by 0.7 raised to the twelfth power
1.5 raised to the eighth power divided by 0.7 raised to the seventh power
2.13
2.115
Mathematics
1 answer:
zaharov [31]2 years ago
3 0

The expression is equivalent to 1.5 raised to the fifth power divided by 0.7 raised to the fourth power, all raised to the third power is; Option A; 1.5 raised to the fifteenth power divided by 0.7 raised to the twelfth power

<h3>How to solve exponents?</h3>

We want to find the expression that is equivalent to 1.5 raised to the fifth power divided by 0.7 raised to the fourth power, all raised to the third power.

The sentence above can be expressed as;

((1.5⁵)/(0.7⁴))³

This can be broken down into;

[1.5^(5 * 3)]/(0.7^(4 * 3))

1.5¹⁵/0.7¹²

Looking at the options, the only correct one is;

Option A; 1.5 raised to the fifteenth power divided by 0.7 raised to the twelfth power

Read more about Exponents at; brainly.com/question/11975096

#SPJ1

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A study found that the mean amount of time cars spent in​ drive-throughs of a certain​ fast-food restaurant was 157.4 seconds. A
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a) The probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds is 5.59%.

b) The probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through is 12.71%

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

A study found that the mean amount of time cars spent in​ drive-throughs of a certain​ fast-food restaurant was 157.4 seconds. This means that \mu = 157.4

Assuming​ drive-through times are normally distributed with a standard deviation of 33 ​seconds. This means that \sigma = 33.

(a) What is the probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds?

This probability is the pvalue of the zscore of X = 105

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

Z = \frac{105 - 157.4}{33}

Z = -1.59

Z = -1.59 has a pvalue of .0559. This means that there is a 5.59% probability that  a randomly selected car will get through the restaurant's drive-through in less than 105 seconds,

(b) What is the probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through?

This is 1 subtracted by the pvalue of the zscore of X = 195.

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

Z = \frac{195 - 157.4}{33}

Z = 1.14

Z = 1.14 has a pvalue of .87286. This means that there is a 87.286% probability that  a randomly selected car will get through the restaurant's drive-through in less than 195 seconds.

So the probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive through is

P = 1 - .87286 = .1271

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