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Leviafan [203]
3 years ago
9

Jeremiah got a job that pays $42K per year. What will Jeremiah’s weekly salary be to the nearest cent?

Mathematics
1 answer:
Katena32 [7]3 years ago
4 0

Answer:

$807.7

Step-by-step explanation:

$42,000÷ 52weeks = 807.69

807.69 rounded to the nearest cent would be 807.7

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How do you write 9,800,100 in exponents and expanded form
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In exponents it could be 3130.51114^2,  213.998224 and 55.95097086
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5 0
3 years ago
The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?​
zimovet [89]

Answer:

0.45 cm

Step-by-step explanation:

Actual length of the micro chip

=  \frac{1}{40}  \times 18 \\  \\  =0.45 \: cm

4 0
3 years ago
Simplify −2xy + 3x − 2xy + 3x and can you explain the answer?
irina1246 [14]

Answer:

 -4xy+6x

Step-by-step explanation:

Given : Expression =  -2xy+3x-2xy+3x

Simplifying the expression :

Step 1. Write the expression =   -2xy+3x-2xy+3x

Step 2.  Add the similar elements:  3x+3x=6x

          =  -2xy-2xy+6x

Step 3. Add the similar elements:  -2xy-2xy=-4xy

           = -4xy+6x

Therefore, -4xy+6x is the simplified form of  -2xy+3x-2xy+3x

8 0
3 years ago
Read 2 more answers
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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