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2 years ago

Can anyone help me with dis

Mathematics

1 answer:

dolphi86 [110]2 years ago

7 0

Answer:

Slope is 2.

Step-by-step explanation:

So to find slope, we need to take the change in the y value, and the change in the x value, and then divide them. This may be confusing, so here is just 3 formulas in 3 seperate steps that we use to find it:

Step 1 - $x_2-x_1=change_.in_.x$

Step 2 - $y_2-y_1=Change_.in_.y$

Step 3 - $\frac{Change_. in_. y}{Change_.in_.x}=slope$

So lets start by doing step one.

All we need is to look for two x values. Lets use 5 and 6:

6-5=1

So 1 is our change of x. This is our 1st step.

Next we can use 2 y values, note that these should be the 2 y values that go wiht the x values above. So this means we have to use -3 and -1:

-1-(-3) = 2

So 2 is our change in y. This is our 2nd step.

Now lets plug this into:

$\frac{Change_. in_. y}{Change_.in_.x}=slope$ :

$\frac{2}{1} = 2$

So our slope is 2. Completing our 3rd and final step.

Hope this helps!

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Step-by-step explanation:

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If the function f(x) translated horizontally to the right by h units, then its image is g(x) = f(x - h)

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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

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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 $\\ P(z>1.56) = 0.0594$ .

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