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Sholpan [36]
2 years ago
15

In a survey of 40 Clemson students, it was found that the mean age (in years) when they would like to get married is 27.4 with a

standard deviation of 6. How many Clemson students would need to be surveyed to estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence?
Mathematics
1 answer:
yKpoI14uk [10]2 years ago
5 0

Answer:

Step-by-step explanation:

Download docx
You might be interested in
Traffic speed: The mean speed for a sample of cars at a certain intersection was kilometers per hour with a standard deviation o
aliina [53]

Answer:

Step-by-step explanation:

Hello!

X₁: speed of a motorcycle at a certain intersection.

n₁= 135

X[bar]₁= 33.99 km/h

S₁= 4.02 km/h

X₂: speed of a car at a certain intersection.

n₂= 42 cars

X[bar]₂= 26.56 km/h

S₂= 2.45 km/h

Assuming

X₁~N(μ₁; σ₁²)

X₂~N(μ₂; σ₂²)

and σ₁² = σ₂²

<em>A 90% confidence interval for the difference between the mean speeds, in kilometers per hour, of motorcycles and cars at this intersection is ________.</em>

The parameter of interest is μ₁-μ₂

(X[bar]₁-X[bar]₂)±t_{n_1+n_2-2} * Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }

t_{n_1+n_2-2;1-\alpha /2}= t_{175; 0.95}= 1.654

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{134*16.1604+41*6.0025}{135+42-2} } = 3.71

[(33.99-26.56) ± 1.654 *(3.71*\sqrt{\frac{1}{135} +\frac{1}{42} })]

[6.345; 8.514]= [6.35; 8.51]km/h

<em>Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15</em>

t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06

[(475.12-321.34) ± 2.485 *(33.06*\sqrt{\frac{1}{12} +\frac{1}{15} })]

[121.96; 185.60]

I hope this helps!

3 0
2 years ago
Find the area of this quadrilateral. Explain or show your strategy.
Verizon [17]

Answer:

24 units

Step-by-step explanation:

1) Separate this quadrilateral into 2 triangles

2) Multiply the height by the width of one of the triangle

- Multiply 8×3

= 24

4 0
1 year ago
Using synthetic division, find (2x4 + 4x3 + 2x2 + 8x + 8) ÷ (x + 2).
Vlad [161]

it is 4 because i used plato and used the answer key

8 0
2 years ago
Read 2 more answers
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C cont
castortr0y [4]

Answer:

k= 80%

Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution  of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid  is added to jar A.

Now jar A contains   4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.

L solute/L solution = 0.5

[4*0.45 L + (2/3)*k/100 L]/(4+2/3)L = 0.5

[1.8 + (2k/300)]/[(12+2)/3] = 0.5

[1.8 + (2k/300)]/[14/3] = 0.5

[1.8 + (2k/300)]= 0.5*(14/3)

(2k/300) = 0.5*(14/3) - 1.8

2k = (0.5*(14/3) - 1.8)*300

k = (0.5*(14/3) - 1.8)*300/2 =80

k= 80%

We also can find k using jar B.

(1/3)k/100 L acid is added  to jar B.

Now jar B contains 5*0.48 L+ (1/3)k/100 L acid, and it has (5+1/3) L of a solution.

L solute/L solution = 0.5

[5*0.48 L+ (1/3)k/100 L ]/(5+1/3)L= 0.5

[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5

This equation also gives k=80%

Check.

We can check at least for jar A.

Jar A has 4L solution and 4*0.45=1.8 L acid.

2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.

(2/3)* 80%= (2/3)*0.8 acid was added from jar C.

Now we have [1.8 +(2/3)*0.8] L acid in jar A.

L solute/L solution =  [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50%  as it is given that jar A has 50% at the end.

7 0
2 years ago
In 10 minutes, Max ran 1.45 km while Avery ran 1.61 km. How much more did<br> Avery run?
SOVA2 [1]

Answer:

0.16km

Step-by-step explanation:

Unless I am missing something it is a simple subtraction problem and the time is not needed. It is simply asking how much farther Avery ran than Max, which is 1.61km - 1.45km which is 0.16km

6 0
1 year ago
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