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Pie
3 years ago
5

Need help for this midterm

Mathematics
1 answer:
Schach [20]3 years ago
6 0

Answer:

Step-by-step explanation:

I can’t read it

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olchik [2.2K]
The coordinates would be (3, 3).

The slope from R to S is given by

m = (0 - 0)/(-4-1) = 0/-5 = 0

The distance from R to S is 5 units straight across.

This means the slope from T to U will be 0, and it will be a horizontal segment.  This means the y-coordinate of U will be 3, since the y-coordinate of T is 3.

The distance from T to U will be 5 as well; -2+5 = 3 for the x-coordinate.

This makes the point (3, 3).
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8 0
3 years ago
The quantities x and y are proportional.<br>pls help plsssss this work is due 20 min plsss helppp
Dahasolnce [82]

Answer: 2.5

Step-by-step explanation:

y = rx

substitute numbers in and solve for r

if

y = 10

x = 4

then

10 = r4

r = 10/4 = 2.5

this works for all the other numbers and always gives r = 2.5

5 0
3 years ago
Hello, may someone plz help me
fenix001 [56]

Answer:

Step-by-step explanation:

this is not high school- but uhm

1) KAT

2)TAK

3) SAT

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4 0
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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