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olga2289 [7]
2 years ago
13

Three consecutive even numbers have a sum where one half of that sum is between 90 and 105.

Mathematics
1 answer:
Darya [45]2 years ago
5 0
n;\ n+2;\ n+4-three\ consecutive\ even\ numbers\\\\90 \ \textless \  \dfrac{1}{2}(n+n+2+n+4) \ \textless \  105\\\\90 \ \textless \  \dfrac{1}{2}(3n+6) \ \textless \  105\ \ \ |multiply\ both\ sides\ by\ 2\\\\180 \ \textless \  3n+6 \ \textless \  210\ \ \ |subtract\ 6\ from\ both\ sides\\\\174 \ \textless \  3n \ \textless \  204\ \ \ |divide\ both\ sides\ by\ 3\\\\58 \ \textless \  n \ \textless \  68\\\\Answer:\\60;\ 62;\ 64\ or\ 62;\ 64;\ 66;\ or\ 64;\ 66;\ 68\ or\ 66;\ 68;\ 70.
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Sin4u=2sin 2u cos 2u​
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Answer:

If you've learnt sin(A+B) = sinAcosB + cosAsinB,

sin(4u)

= sin(2u+2u)

= sin(2u)cos(2u) + cos(2u)sin(2u)

= 2 sin(2u) cos(2u).

4 0
2 years ago
How much work must be done on a 1000-kg car to increase its speed from 1 m/s
yarga [219]

Answer:

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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
2 years ago
Pls help ASAP so I can't fail and show proof why you pick that answer!
cluponka [151]

Answer:

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Step-by-step explanation:

7 0
2 years ago
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Answer:

3/4

Step-by-step explanation:

7 0
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