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dimaraw [331]
3 years ago
11

Guys i really need your help i have this awesome friend who is cool sometimes but she calls me names and pretends to like me bot

tom line is: is she really my friend
#cryingmyeyesoutrn
Mathematics
2 answers:
garik1379 [7]3 years ago
8 0

If she calls you name it isn’t very nice and If she pretends to like you than she ain’t a very good friend

Lelechka [254]3 years ago
7 0

Answer:

No. If your being treated like trash and are constantly being called names, talk to your parents or someone, they can help. It's up to you to decide. There are many websites with great advice.

Step-by-step explanation:

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Kim is having a pizza party. Two large pizzas serve 9 people. How many large pizzas should she order to serve 36 guests at the p
Angelina_Jolie [31]
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8 large pizzas for 36
(Me personally would eat 8 all by myself)
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Rudiy27
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What is 23% of 256? please help
Anastasy [175]
So first you need to turn the percent into a decimal. So 23% would be 0.23 because you divide 23 by 100 to turn it into a decimal. Now you multiply 0.32 and 256. So:

0.23 *256=58.88 

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Step-by-step explanation:

5 0
2 years ago
Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a
tigry1 [53]

Answer:

SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58

ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073

\bar X - ME = 237- 14.073 = 222.927

\bar X + ME = 237+ 14.073 = 251.073

n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis:\mu \geq 247  

Alternative hypothesis:\mu  

z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165  

p_v =P(z  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

\bar X =237 represent the sample mean

\sigma = 47 represent the population deviation

n =30 represent the sample size selected

\mu_0 = 247 represent the value that we want to test.

The standard error for this case is given by:

SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58

For the 90% confidence the value of the significance is given by \alpha=1-0.9 = 0.1 and \alpha/2 = 0.05 so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

z_{\alpha/2}= 1.64

And the margin of error would be:

ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073

The confidence interval for this case would be given by:

\bar X - ME = 237- 14.073 = 222.927

\bar X + ME = 237+ 14.073 = 251.073

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

Replacing into formula (b) we got:

n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:  

Null hypothesis:\mu \geq 247  

Alternative hypothesis:\mu  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165  

P-value  

Since is a left tailed test the p value would be:  

p_v =P(z  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

5 0
3 years ago
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