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3 years ago

What number is 25% of 180

2 answers:

7 0

$p\%=\frac{p}{100}\\\\25\%=\frac{25}{100}=\frac{25:25}{100:25}=\frac{1}{4}\\\\25\%\ of\ 180=\frac{1}{4}\ of\ 180=\frac{1}{4}\cdot180=\frac{180}{4}=\boxed{45}$

TiliK225 [7]3 years ago

4 0

'what number' indicates an unknown value we can represent with 'x'

'is' indicates an equal sign.

25% = 0.25

'of' indicates multiplication.

x = 0.25 * 180

x = 45

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What is the answer to the equation 14.95x6.5?

NeTakaya

Answer:

97,175

Step-by-step explanation:

I don't think It's necessary, but I am gonna do it anyways.

15*6=90

15*7=105

105 - 90 =15

15/2 =7,5

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15*6,5= 97,5

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2 years ago

What is the solution for -10 < x - 9?

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ms evans went to the store to buy some cupcakes. she brought 129 cupcakes. she wants to put 5 cupcakes in each box.how many boxe

NeTakaya

129 divided by 5 is equal to 25 with a remainder of 4.

This means that she will need a total of 26 boxes, but only 25 of them will have 5 cupcakes in them. The last one will have 4 cupcakes.

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Which of the Following statements is NOT true?

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Answer:

the statement which is not true is

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3 years ago

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an

solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. (a) What is the minimum pregnancy length that can be in the top 11% of pregnancy lengths? (b) What is the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths?

Solution:

The random variable X can be defined as the pregnancy length in days.

Then, from the provided information $X\sim N(\mu=268, \sigma^{2}=12^{2})$ .

(a)

The minimum pregnancy length that can be in the top 11% of pregnancy lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

⇒ z = 1.23

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283$

Thus, the minimum pregnancy length that can be in the top 11% of pregnancy lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom 5% of pregnancy lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ z = -1.645

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248$

Thus, the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths is 248 days.

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3 years ago