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Alexeev081 [22]
3 years ago
7

What is the value of x? 3 6 27 72

Mathematics
1 answer:
DerKrebs [107]3 years ago
5 0
I have no freakin idea of this
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The rectangles shown below are similar figures. Use the properties of similar figures to solve for the value of s.
Sergeu [11.5K]

Answer:

s = 3

Step-by-step explanation:

we know the rectangles are similar

2 × __ = 16

s × __ = 24

so,

16÷2 = 8

2 × 8 =16

s × 8 =24

then,

24 ÷ 8 = s

3 = s

2 × 8 = 16

3 × 8 =24

hoped i helped :)

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2 years ago
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Multiply 5,5,5,5 4 times which gives me 625
k0ka [10]

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Confusion

Step-by-step explanation:

what the heck are you asking?

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3. The photo below is being enlarged to form
Umnica [9.8K]

Answer:

Not enough information

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2 years ago
Simplify this answer pls
ipn [44]

Answer:

D

Step-by-step explanation:

when it's a power of the power we multiply the powers to get a single value for the power.

(6^(1/4))^4=6^(4*(1/4)) (4*(1/4)=1)

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7 0
3 years ago
"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.

4 0
3 years ago
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