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viktelen [127]
3 years ago
13

Need help with 56 plz!

Mathematics
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer: No idea, sorry. I'm sure you could look it up tho


Step-by-step explanation:


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Find the value of x please help I give brainly<br> 20 points
fiasKO [112]

Answer:

x=148

Step-by-step explanation:

First, find the missing angle by subtracting 100+48 from 180 (all of the angles in a triangle add up to 180). Then, subtract that from 180 (the angles are a linear pair, so they are supplementry, so they add up to 180) x+32=180.

x=148.

(お役に立てれば!)

7 0
2 years ago
Find the value of x.<br> 8:5 = 1/x:1
tatiyna

Answer:x=61

Step-by-step explanation​

 

856−5x3{131}=2

8(x−9)+5(5−18)+1(15−6x)=2

8x−72−65+15−6x=2

2x=122

x=61

sorry if i'm wrong

5 0
1 year ago
the equation of line t is 6x+5y=3, and the equation of line q is 5x-6y=0. Which statement about the two lines is true?
nekit [7.7K]
The correct answer would be: 6x+5y=3. I am not entirely sure, but that's what I believe is the correct answer.-Hope this helped!
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
What is the nearest thousand to 700,000
emmasim [6.3K]
It would stay 700,000
3 0
3 years ago
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