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kaheart [24]
3 years ago
6

A car travels 30 1/5 miles in 2/3 of an hour

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
8 0
Send me the pics of y’all in my brothers friends and family friends
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If one zero of the polynomial (a^2+9)x^2+13x+6a is the reciprocal of the other, find the value of a.​
mihalych1998 [28]

Answer: a^2x^2+9x^2+13x+6a

Step-by-step explanation:

(a^2+9) x^2+ 13x +6a

x^2(a^2+9)+13x+6a

Expand:x^2(a^2+9): a^2 x^2+9^2

x^2 a^2 +x^2*9

a^2 x^2+9x^2

  a^2 x^2+9x^2+13x+6a

5 0
2 years ago
What is the surface area of a sphere with a diameter of 8 mm?<br> Use 3.14 for π .
Rudik [331]

Answer:

Step-by-step explanation:

804.25mm

8 0
3 years ago
Which set of measures would give a clear conclusion about the data. Help please.
adell [148]
<h2>We need more details for this question. We may not understand it because we could not see the data.</h2>

8 0
2 years ago
Am I correct? If not please explain why!
Kipish [7]

yes, you are correct. every input corresponds to only one output

6 0
2 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
2 years ago
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