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4 years ago

Use the confidence level and sample data to find a confidence interval

Mathematics

1 answer:

stepan [7]4 years ago

4 0

Answer:

The confidence interval at at 99% level of confidence is 93.7 ≤ μ ≤ 96.9.

Step-by-step explanation:

Step 1:

We must first determine the z-value at a confidence level of 99%.

Therefore,

99% = 100%(1 - 0.01)

Thus,

α = 0.01

Therefore, the z-value will be

z_(α/2) = z_(0.01/2) = z_0.005 = 2.58

(The z-value is read-off from the z table from the standard normal probabilities.)

Step 2:

We can now write the confidence interval:

X - z_(α/2) [s/√(n)] ≤ μ ≤ X + z_(α/2) [s/√(n)]

95.3 - 2.58(6.5/√(104)) ≤ μ ≤ 95.3 + 2.58(6.5/√(104))

93.7 ≤ μ ≤ 96.9

Therefore, confidence interval is 93.7 ≤ μ ≤ 96.9 which means that we are 99% confident that the true mean population lies is at least 93.7 and at most 96.9.

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(cot^2x - 1)/(csc^2x) = cos2x

Alex787 [66]

Answer:

Step-by-step explanation:

The idea here is to get the left side simplified down so it is the same as the right side. Consequently, there are 3 identities for cos(2x):

$cos(2x)=cos^2x-sin^2x$ ,

$cos(2x)=1-2sin^2x$ , and

$cos(2x)=2cos^2x-1$

We begin by rewriting the left side in terms of sin and cos, since all the identities deal with sines and cosines and no cotangents or cosecants. Rewriting gives you:

$\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

Notice I also wrote the 1 in terms of sin^2(x).

Now we will put the numerator of the bigger fraction over the common denominator:

$\frac{\frac{cos^2x-sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

The rule is bring up the lower fraction and flip it to multiply, so that will give us:

$\frac{cos^2x-sin^2x}{sin^2x} *\frac{sin^2x}{1}$

And canceling out the sin^2 x leaves us with just

$cos^2x-sin^2x$ which is one of our identities.

5 0

3 years ago

HELP

Ugo [173]

Answer:

C. 22 M

Step-by-step explanation:

7 0

3 years ago

HELP ASAP What is the coefficient of the term 6/7xy ?

yawa3891 [41]

Answer: The coefficient of the 6/7xy is 6/7.

7 0

3 years ago

Read 2 more answers

What is 4 times as much as 20.075? show work

choli [55]

Answer:

80.3

Step-by-step explanation:

20.075 x 4 = 80.3

Hope that helps!

5 0

2 years ago

Brandon is on one side of a river that is 50 m wide and wants to reach a point 300 m downstream on the opposite side as quickly

AlexFokin [52]

Let P be Brandon's starting point and Q be the point directly across the river from P.
Now let R be the point where Brandon swims to on the opposite shore, and let 
QR = x. Then he will swim a distance of sqrt(50^2 + x^2) meters and then run 
a distance of (300 - x) meters. Since time = distance/speed, the time of travel T is 

T = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x). Now differentiate with respect to x: 

dT/dx = (1/4)*(2500 + x^2)^(-1/2) *(2x) - (1/6). Now to find the critical points set 
dT/dx = 0, which will be the case when 

(x/2) / sqrt(2500 + x^2) = 1/6 ----> 

3x = sqrt(2500 + x^2) ----> 

9x^2 = 2500 + x^2 ----> 8x^2 = 2500 ---> x^2 = 625/2 ---> x = (25/2)*sqrt(2) m, 

which is about 17.7 m downstream from Q. 

Now d/dx(dT/dx) = 1250/(2500 + x^2) > 0 for x = 17.7, so by the second derivative 
test the time of travel, T, is minimized at x = (25/2)*sqrt(2) m. So to find the 
minimum travel time just plug this value of x into to equation for T: 

T(x) = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x) ----> 

T((25/2)*sqrt(2)) = (1/2)*(sqrt(2500 + (625/2)) + (1/6)*(300 - (25/2)*sqrt(2)) = 73.57 s.



mind blown

8 0

3 years ago