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3 years ago

I need u to show me work.about 4 divided by 1/4

2 answers:

n200080 [17]3 years ago

6 0

$\bf 4 \div \cfrac{1}{\underline{4}}\implies \cfrac{4}{1} \div \cfrac{1}{\underline{4}}\implies \cfrac{4}{1} \cdot \cfrac{\underline{4}}{1}\implies 16$

KengaRu [80]3 years ago

4 0

4 divided by 1/4
4 times 4/1
4 times 4 equals 16

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SIMPLIFY the expression! Please!

Levart [38]

1+3^2 = 1+9 = 10

10/5 = 2

2-6+2 = -4+2

answer is -2

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3 years ago

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Llana [10]

Answer:

3x was subtracted from the left side, but 3x was subtracted from the right side. The Subtract Property of Equality states that you can subtract the same number from each side and the equation will remain true. But 3x and 3 are not the same number (unless x is 1).

x = -8/5

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

$(7x+3x)+(21)=5$
$10x + 21 = 5$

Step 2: Subtract 12 from both sides.

$10x + 21 - 21 = 5 - 21$
$10x = -16$

Step 3: Divide both sides by 10.

$\frac{10x}{10} = \frac{-16}{10}$
$x = -\frac{8}{5}$

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3 years ago

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kvasek [131]

Answer:

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Step-by-step explanation:

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3 years ago

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Stels [109]

Answer:

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2 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.

8 0

3 years ago