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gayaneshka [121]
3 years ago
10

A social psychologist interested in cultural differences compared women of two ethnic groups on a Role Approval Index on which h

igh scores indicated high degrees of approval of one's own social role. The results were as follows.
Ethnic Group A: N=15, M=55, S2=6.5

Ethnic Group B: N=23, M=51, S2=4.5

If the standard deviation of the distribution of the difference between means is .76, what is the t score?

a. 5.26 b. -10.53 c. -8.00 d. 4.18
Mathematics
1 answer:
Levart [38]3 years ago
3 0

Answer:

a. 5.26

Step-by-step explanation:

As, the standard deviation of the distribution of the difference between mean is 0.76 so the t score will be simply calculated by dividing the difference of means by standard deviation of the distribution of the difference between mean as hypothesized difference is zero.

t score=Xbar A- Xbar B/Sd of difference of means

t score=55-51/0.76

t score=4/0.76=5.26

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Four less than the quotient of a number and 4 is 14. What is the number? Let n be the unknown number.
Maru [420]
So the equation is n/4-4=14
then you +4 to both sides 
n/4-4=14 so now the equation is n/4=18 then you multiply 18 by 4 and get 72
   +4    +4
then to check your work
72/4-4=14
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Hope I helped :)

8 0
3 years ago
An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W. How far is the plane f
Ganezh [65]

Answer:

299.99 miles

Step-by-step explanation:

Since the plane traveled due west,

The total angle is 49.17 + 90

Represent that with θ

θ = 49.17 + 90

θ = 139.17.

Represent the sides as

A = 170

B = 150

C = unknown

Since, θ is opposite side C, side C can be calculated using cosine formula as;

C² = A² + B² - 2ABCosθ

Substitute values for A, B and θ

C² = 150² + 170² - 2 * 150 * 170 * Cos 139.17

C² = 22500 + 28900 - 51000 * Cos 139.17

C² = 51400 - 51000 (−0.7567)

C² = 51400 + 38,591.7

C² = 89,991.7

Take Square Root of both sides

C = 299.9861663477167

C = 299.99 miles (Approximated)

Hence, the distance between the plane and the airport is 299.99 miles

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