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xxMikexx [17]
3 years ago
10

Which of the sets of terms below does NOT include like terms? *

Mathematics
1 answer:
antoniya [11.8K]3 years ago
6 0
6x, 6y, 6z. Different variables means they aren’t like terms. Also love ur pfp hehe :)
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A bag contains 9 red marbles, 18 orange marbles, 2 yellow marbles, and 5 purple marbles. A marble is drawn at random from the ba
vlada-n [284]

Answer:

8/17

Step-by-step explanation:

total balls= 9 + 18 + 2 + 5 = 34

balls drawn should NOT be Orange.

so, 34 - 18 (as there are 18 orange balls)

16 balls

probability= 16/34 = 8/17

answer = 8/17

7 0
3 years ago
NO LINKS PLEASE HELP NOW COMPLETELY ACCURATE ANSWER IMMEDIATELY PLEASE ANSWER I WILL GIVE BRAINLIEST
aivan3 [116]

Answer:

got em

Step-by-step explanation:

8 0
3 years ago
A New York Times/CBS News Poll asked a random sample of U.S. adults the question, "Do you favor an amendment to the Constitution
AnnZ [28]

Answer:

p_v =P(z>-0.436)=0.669  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is NOT significantly higher than 2/3.

We got the same conclusion just looking the confidence interval since our interval contains the value 2/3=0.667 we have enough evidence to fail to reject the null hypothesis on this case.  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

For this case we can get an estimation for \hat p like this:

\hat p= \frac{0.63+0.69}{2}=0.66

And the margin of error would be:

ME= \frac{0.69-0.63}{2}=0.03

And we can estimate the standard error since ME= z SE

SE = \frac{0.03}{1.96}=0.0153

We need to conduct a hypothesis in order to test the claim that the proportion is higher than 2/3 or not:  

Null hypothesis:p \leq \frac{2}{3}  

Alternative hypothesis:p > \frac{2}{3}  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.66 -\frac{2}{3}}{0.0153}=-0.436  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is \alpha=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(z>-0.436)=0.669  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is NOT significantly higher than 2/3.

We got the same conclusion just looking the confidence interval since our interval contains the value 2/3=0.667 we have enough evidence to fail to reject the null hypothesis on this case.  

8 0
4 years ago
Jina has scored 79, 85, 89 and 68 on her previous four tests. What score does she need on her next test so that her mean is 80?
poizon [28]

Answer:

79

Step-by-step explanation:

79, 85, 89, 68 have a mean of 80.25 according to <em><u>calculator.net</u></em>. the number itself Is close to the number you want, in this case 80, i input 80, but it gave me a score of 80.2, so i tried again with 81, it gave me a score of 80.4, so the third attempt was successful with 79 being the input and 80 the output

calculator.net is approved on my school computer so there is a chance it is on yours as well.

4 0
3 years ago
Help <br> "It's too short. Write at least 20 characters to get a great answer."
Shalnov [3]

Answer:

Because for the first problem you would have to simplify 8 divided by 2 into 4  in which  12 - 4 = 8 then you would have to simplify 12 - 4 to 8 which 8 = 8

Because  for the second problem you would have to simplify 12 - 8 to 4 which  4 divided by 2 = 2 then you would have to simplify 4 divided by 2 to 2 which 2 = 2

There are parentheses which if there weren't you would get a different answer

Hope I helped

Sorry if I didnt

8 0
3 years ago
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