2x + 5y - 28 -2x - 5y = -18 Elimination

Shelbi wants to run one lap around the track as fast as her friend Annabelle, who can run one lap in 1 minute and 10 seconds. Sh

Mamont248 [21]

23 rrrrrrrrrrrrrrrrrrrrrrrrrrrrr

4 0

3 years ago

Read 2 more answers

What is the square root of 14

harina [27]

The question is asking " What is the square root of 14?
3.74
But, if you need to round to the nearest tenth, your answer should come out to be 3.7

8 0

3 years ago

Laci needs to put some logs in the back of her pickup truck. The dimensions of the truck bed are 8 feet by 5 .5 feet. What is th

Mazyrski [523]

Since the truck is 8 feet long then the longest log able to fit in the truck should be 8 feet as well. However talking reality, we might want to shorten it to make it fit better. Hope this helps!

3 0

3 years ago

Suppose that a dog, a goat, and a bag of tin cans are to be transported across a river in a ferry that can carry only one of the

Bingel [31]

Answer:

The task is explained in detail below.

Step-by-step explanation:

We know that a dog, a goat, and a bag of tin cans are to be transported across a river in a ferry that can carry only one of these three items at once (along with a ferry driver).

The driver will first transport the goat to the other shore and then return empty. Then they will take a bag of tin and transport it to the other shore, and then they will return with the goat to their original place.

Then he will leave the goat there and take a dog and carry it to the other shore and then return empty.

And eventually they will take the goat and transport it to the other shore.

4 0

3 years ago

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 distribution of sample means. 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 normally 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 normal standard distribution, 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 distribution of sample means. 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 standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table 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 complement probability 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