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Nimfa-mama [501]

3 years ago

5

A chicken’s mass at birth is 0.40 kilograms. The chicken gains approximately .44 kilograms each week. After how many weeks is th

e chicken’s mass 9.2 kilograms?

1 answer:

tresset_1 [31]3 years ago

5 0

About 20 weeks because 9.2/.44 is around 20

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Decimals from 7.0 to 8.4 with an interval of 0.2 between each pair of decimals

Vitek1552 [10]

Decimals from 7.0 to 8.4 with an interval of 0.2 between each pair of decimals

We start with 7.0 and keep adding 0.2 till we get 8.4

7.0 + 0.2 = 7.2

7.2 + 0.2 = 7.4

7.4 + 0.2 = 7.6

7.6 + 0.2 = 7.8

7.8 + 0.2 = 8.0

8.0 + 0.2 = 8.2

8.2 + 0.2 = 8.4

So, the decimals from 7.0 to 8.4 with an interval of 0.2 are

7.2, 7.4, 7.6, 7.8, 8.0, 8.2

3 0

3 years ago

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You're given two side lengths of 3 centimeters and 5 centimeters. Which measurement can you use for the length of the third side

igomit [66]

Answer: To find the third side of a triangle (there are two answers you could get), you need to use the Pythagorean formula: a^2+b^2=c^2. First answer, 3^2+5^2=c^2. Solve for c. c=5.83. The second answer, 3^2+b^2=5^2. Solve for b. b=4.

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5 0

3 years ago

ONLY ANSWER IF YOU KNOW THE ANSWER JUST FOR POINTS. Thank you! :)

Ksivusya [100]

Since triangle DEF = triangle JKL, m<D = m<J, m<E = m<K, m<F = m<L.
m<F = m<L = 90 degrees
m<K = m<E = 5(m<D)
but m<E + m<D = 90 degrees [right angled triangle]
5(m<D) + m<D = 90 degrees
6(m<D) = 90 degrees
m<D = 90 / 6 = 15 degrees.

5 0

3 years ago

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The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any <em>cumulative</em> <em>standard normal table</em> available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

3 years ago

Can someone please help me

borishaifa [10]

Add up all the sides and you get 10

4 0

3 years ago

Read 2 more answers