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2 years ago

12

Diego has run 4/5 of a race distance. He has run 7 2/3

1 answer:

lianna [129]2 years ago

8 0

Your question is incomplete, but I am guessing that it reads something like:

“Diego has run 4/5 of a race distance. He has run 7 2/3 miles. How long is the race? Or “how many miles does he have left?”

How long is the race? = 9 7/12 miles

4/5x = 7 2/3
4/5x = 23/3
x = 23/3 • 5/4
x = 115/12
x = 9.58333
x = 9 7/12 miles

How many miles does he have left? = 1 1/12 miles

1/5 • 115/12 = 115/60 = 1.916666
= 1 55/60 = 1 11/12 miles

Let me know if I didn’t guess your question correctly and I will try to help more.

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2 years ago

Solve for x: −2(x + 3) = −4(x + 1) − 4

Stolb23 [73]

Hello! My name is Chris and I’ll be helping you with this problem.

Date: 9/27/20 Time:

Answer:

x = -1

Explanation:

Step 1: Simplify both sides of the equation.

−2(x+3)=−4(x+1)−4

(−2)(x)+(−2)(3)=(−4)(x)+(−4)(1)+−4(Distribute)

−2x+−6=−4x+−4+−4

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−2x−6=−4x−8

Step 2: Add 4x to both sides.

−2x−6+4x=−4x−8+4x

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Step 3: Add 6 to both sides.

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Step 4: Divide both sides by 2.

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/2 /2

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I hope this helped answer your question! Have a great rest of your day!

Furthermore,

Chris

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2 years ago

Read 2 more answers

"A laboratory tested 73 chicken eggs and found that the mean amount of cholesterol was 230 milligrams with LaTeX: \sigmaσ = 17.4

Scrat [10]

Answer:

The 95% confidence interval for the true mean cholesterol content, μ, of all such eggs is between 226.01 and 233.99 milligrams.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{17.4}{\sqrt{73}} = 3.99$

The lower end of the interval is the sample mean subtracted by M. So it is 230 - 3.99 = 226.01

The upper end of the interval is the sample mean added to M. So it is 230 + 3.99 = 233.99.

The 95% confidence interval for the true mean cholesterol content, μ, of all such eggs is between 226.01 and 233.99 milligrams.

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