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pav-90 [236]
3 years ago
8

Can someone help me? before its late

Mathematics
1 answer:
Akimi4 [234]3 years ago
7 0
1. X+18
2. Y-5
3. x+7=y
4. Y-10
5. 7-x
16. r-s
17. R+25
18. z-12
19. m+9
20. Y-11
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Ismael has a set of 5 solar panels and Ben has a set of 6 smaller solar panels. Each set of panels produce the same amount of en
eimsori [14]

9514 1404 393

Answer:

  Ben's

Step-by-step explanation:

If E is the amount of energy that each full set of panels produces, then ...

  3 of Ismael's 5 panels produce 3/5E

  4 of Ben's 6 panels produce 4/6E

We can compare these fractions when they have a common denominator.

  3/5E = 18/30E . . . . energy from Ismael's panels

  4/6E = 20/30E . . .  energy from Ben's panels

  18/30 < 20/30 . . . . so Ben's panels are producing more energy

5 0
3 years ago
A newspaper poll asked respondents if they trusted "eco friendly" labels on cleaning products. Out of 1000 adults surveyed, 498
Paul [167]

Answer:

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 respondents that trust these labels is not significanlty less than 0.5 or 50%.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=498 represent the adults that trust these labels

\hat p=\frac{498}{1000}=0.498 estimated proportion of respondents that trust these labels

p_o=0.5 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of respondents that trust these labels is at least 50%:  

Null hypothesis:p\geq 0.5  

Alternative hypothesis:p < 0.5  

When we conduct a proportion test we need to use the z statisitc, 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.

3) Calculate the statistic  

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

z=\frac{0.498 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=-0.126  

4) 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 is not provided but we can assume it as \alpha=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(Z  

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 respondents that trust these labels is not significanlty less than 0.5 or 50%.  

7 0
3 years ago
Jim wanted to find out what the audience thought about the debate. After the event, Jim stood at the exit to survey every fifth
balu736 [363]

Answer: D. This was a random sample. It may have included anyone in attendance.

Step-by-step explanation:

The options are:

A. This was a biased sample. Jim should interview all in attendance.

B. This was a census. Any guest may have participated.

C. This was a random sample. It may not have included anyone in attendance.

D. This was a random sample. It may have included anyone in attendance.

A random sampling is simply referred to as a subset of individuals that are picked from a larger set of individuals.

With regards to the question, Jim wanted to find out what the audience thought about the debate and after the event, he stood at the exit to survey every fifth guest.

This means that it was a random sampling and anyone could have been picked, the sampling wasn't bias.

7 0
3 years ago
Sara bought a 25-ounce jar of strawberry jam for $4.50. What is the unit price?
fgiga [73]
$0.18 is the answer. How I got it was you do $4.50 over 25 ounces. Then divide the $4.50 by 25. You should get 0.18
7 0
3 years ago
What is the mean absolute deviation of the following set of data 4 6 2 8
babymother [125]

Answer:

srry i really need points

Step-by-step explanation:

6 0
2 years ago
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