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saul85 [17]
2 years ago
14

Find the area of a circle with a diameter of 16.

Mathematics
1 answer:
tangare [24]2 years ago
8 0

Answer:

Step-by-step explanation:

Area of circle:

<em>area = π · r · r</em>

<em>Radius=  </em>\frac{16}{2}= 8

3.14\times { 8  }^{ 2  } = 200.96 cm^2\\

<em></em>

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Solve the inequality <br> 2x-1/x &gt; 1
Grace [21]

Answer:

− 1 /2 < x < 0

or  x > 1

Step-by-step explanation:

Solve the inequality by finding the roots and creating test intervals.

Glad I could help!!

8 0
3 years ago
Read 2 more answers
The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability
amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

Z = \frac{X - \mu}{\sigma}

By the Central limit theorem

Z = \frac{X - \mu}{s}

Z = \frac{70.5 - 69}{1.97565}

Z = 0.76

Z = 0.76 has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0
3 years ago
Please help me ASAP
expeople1 [14]
Your equation would be:

<span>y=15*0.06^x</span>

7 0
3 years ago
PLEASE NO LINKS OR SITES!!!! JUST ANSWERS. THANK YOU!
lutik1710 [3]

Answer:

F=-9

Hope This Helps!

6 0
2 years ago
How to make an equation with numbers 0.25, 7.00, and 27
shepuryov [24]
There are many ways to do this.

One way could be 0.25x+7.00=27
8 0
3 years ago
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