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aev [14]
3 years ago
14

Nina earns $15 for each yard she mows the table shows her earnings for 0, 1, 2 and 3 yards mowed how much will Nina earn if she

mows 6 yards?

Mathematics
1 answer:
Paha777 [63]3 years ago
8 0
Hello there!

According to the chart, you can see that for every lawn Nina mows, she earns $15. So it's increasing by 15 per yard.

Let's see if this will help:

1 yard= $15
6 yards = ???

I don't want to give you the answer right away, since I want you to learn how to do it. :)

I hope my hint helped!!

(If you're really stuck, take 6 times 15)!


Hope this helped!!

-----------


DISCLAIMER: I am not a professional tutor or have any professional background in your subject. Please do not copy my work down, as that will only make things harder for you in the long run. Take the time to really understand this, and it'll make future problems easier. I am human, and may make mistakes, despite my best efforts. Again, I possess no professional background in your subject, so anything you do with my help will be your responsibility. Thank you for reading this, and have a wonderful day/night!
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A market surveyor wishes to know how many energy drinks teenagers drink each week. They want to construct a 80% confidence inter
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Answer:

The 80% confidence interval for the mean

(4.6199 , 4.7801)

Step-by-step explanation:

<u>Explanation</u>:-

Assuming that the population standard deviation for the number of energy drinks consumed each week is 1

Given the Population standard deviation 'σ' = 1

The study found that for a sample of 256 teenagers the mean number of energy drinks consumed per week is 4.7

given sample size 'n' = 256

mean of the sample 'x⁻' = 4.7

<u>confidence interval for the mean</u>

The 80% confidence interval for the mean is determined by

(x^{-} -Z_{\alpha } \frac{S.D}{\sqrt{n} } , x^{-} + Z_{\alpha }\frac{S.D}{\sqrt{n} } )

the z-score of 80% level of significance = 1.282

(4.7 - 1.282\frac{1}{\sqrt{256} } , 4.7 + 1.282\frac{1}{\sqrt{256} } )

(4.7 - 0.0801 , 4.7 +0.0801)

(4.6199 , 4.7801)

<u>Conclusion</u>:-

The 80% confidence interval for the mean

(4.6199 , 4.7801)

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