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BabaBlast [244]
4 years ago
5

Nancy sold 18 fewer boxes than Lucas. How many boxes did Lucas sell?

Mathematics
1 answer:
Jet001 [13]4 years ago
6 0

This question is incomplete. Here is the complete question:

Nancy have sold 19 box in total. Nancy sold 18 fewer boxes than Lucas,

How many boxes did Lucas sell?

Answer:

The correct answer is 37.

Step-by-step explanation:

First we must analyze the data we have.

We know that the total number of boxes sold by Nancy is 19.

We know that Nancy sold 18 boxes less than Lucas.

But we don't know how many boxes Lucas sold, so we'll assign an x.

Now, knowing that Nancy sold 18 boxes less than Lucas, and that 19 is the total, the equation will be as follows:

x-18 = 19

The next thing we should do is try to make the term x remain alone, therefore we must move the -18, always remembering that a sum or a subtraction <u>passes to the other side with the opposite sign.</u> Then it will be:

x = 19 + 18

Then we solve the operation:

x = 37

In this way the variable x already has a result that is 37.

Therefore the correct answer is that Lucas sold 37 boxes.

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What is the value of -91/5 - 10 + 2/5+ 2(14 1/2-7)
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Step-by-step explanation:

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3 years ago
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $10,
Andreyy89

Answer:

1) the planning value for the population standard deviation is 10,000

2)

a) Margin of error E = 500, n = 1536.64 ≈ 1537

b) Margin of error E = 200, n = 9604

c) Margin of error E = 100, n = 38416

3)

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

Step-by-step explanation:

Given the data in the question;

1) Planning Value for the population standard deviation will be;

⇒ ( 50,000 - 10,000 ) / 4

= 40,000 / 4

σ = 10,000

Hence, the planning value for the population standard deviation is 10,000

2) how large a sample should be taken if the desired margin of error is;

we know that, n = [ (z_{\alpha /2 × σ ) / E ]²

given that confidence level = 95%, so z_{\alpha /2  = 1.96

Now,

a) Margin of error E = 500

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 500 ]²

n = [ 19600 / 500 ]²

n = 1536.64 ≈ 1537

b) Margin of error E = 200

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 200 ]²

n = [ 19600 / 200 ]²

n = 9604

c)  Margin of error E = 100

n = [ (z_{\alpha /2 × σ ) / E ]²

n = [ ( 1.96 × 10000 ) / 100 ]²

n = [ 19600 / 100 ]²

n = 38416

3) Would you recommend trying to obtain the $100 margin of error?

As we can see, sample size corresponding to margin of error of $100 is too large and may not be feasible.

Hence, I will not recommend trying to obtain the $100 margin of error in the present case.

7 0
3 years ago
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