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Mama L [17]
3 years ago
6

Please help!!!! Giving brainliest and extra points!!!! (Click on the photo)

Mathematics
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

i dont understand the question much cuz its only told berries so i think is both rasberry and strawberry

Step-by-step explanation:

if so here my solve: Both berries are 38% of the graph, we have equation 38%×290 = 110,2

hmmmm maybe the ans is D. 110

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Please help me asappp (will give brainiest!)
ryzh [129]

Answer:

TanФ=-1.793

Step-by-step explanation:

Sin^{2} (A) + Cos^{2} (A) = 1

Sin(A)=\sqrt{1-Cos^{2} (A)}

Tan(A) = \frac{Sin(A)}{Cos(A)}

Tan(A) =\frac{\sqrt{1-Cos^{2}(A) } }{Cos(A)} \\Tan(A) =\frac{\sqrt{1-(-.487)^{2} } }{-.487}=-1.793433202

3 0
3 years ago
Nick can run the 440 yard dash in 55 seconds, and Jack can run it in 88 seconds. How great a handicap must Nick give Jack for th
Zepler [3.9K]
Jack --> 440 = rate * 88Jack --> 5 yrds/sex = rateJack in 55 seconds --> distance = 5 * 55 = 275 yrds440 - 275 = 165 yards handicap.
8 0
3 years ago
Tanya's father travels every week for business. He has traveled around the
Nikitich [7]

The correct answer would be: 4.6 x 10^5

8 0
2 years ago
The next number in the pattern 1 5 25 125 625
chubhunter [2.5K]
5⁰ = 1
5¹ = 5
5² = 25
5³ = 125
5⁴ = 625
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5 0
4 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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