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3 years ago

What is 120% of 180?

Mathematics

2 answers:

Advocard [28]3 years ago

6 0

The answer is 66.7. I hope this is right

Tpy6a [65]3 years ago

6 0

Step-by-step explanation:

$\tt{120 \: \% \: of \: 180}$

Convert % to fraction. For that , Divide 120 by 100 :

⇢ $\tt{ \frac{120}{100} \times 180 }$

Calculate

⇢ $\tt{216}$

$\red{ \boxed{ \boxed{ \tt{Our \: final \: answer : \boxed{ \underline{ \tt{216}}}}}}}$

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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X + y = -15. Subtract x from both sides.

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Select all the exspressions that are all equivalent to 20 percent of 150

Dmitriy789 [7]

Note: It seems you may have unintentionally missed adding the answer choices. Thus, I am solving your question in general to give you the idea of how the percentage works, which anyways would solve your query.

Answer:

Please check the explanation.

Step-by-step explanation:

Given that we have to determine the expressions which are equivalent to 20 percent of 150.

First, we need to determine what actually 20 percent of 150 really brings.

i.e

20% of 150 = 20/100 × 150

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20% of 150 = 30

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2 years ago

What are the solutions to the quadratic equation below? X^2-8x+16

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(₋4)²

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3 years ago

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Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

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

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

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