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dmitriy555 [2]
3 years ago
9

A student has taken three math tests so far this semester His scores for the first three tests were 76, 80, and 84

Mathematics
1 answer:
Charra [1.4K]3 years ago
8 0

Answer:

A=96%         B=86%

Step-by-step explanation:

brainlist will be welcomed :))

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Which of the following would be a better buy than purchasing 4 mangoes for $16?
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Each of a group of 20 intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut str
sergeinik [125]

Answer:

Find answers below

Step-by-step explanation:

H0: P <= 0.5

Ha: P > 0.5

, the number who prefer gut strings is <= a number or the test tends towards the  left-tailed.

{0,1,2,3,4,5} ;  

{15,16,17,18,19,20} is a right-tailed test and not appropriate for H0:

{ 0,1,2,3,17,18,19,20} is two-tailed and not appropriate for H0:

b)

Does the region specify a level .05 test? No

 

P = proportion who prefer gut strings to nylon

P = X /20

Assume alpha = 0.05

z(alpha) = -1.645

Reject if (x/20 - 0.5) / sqrt[ (0.5)(0.5)/20 ] < -1.645

Reject if (x/20 - 0.5) <  < (-1.645) sqrt ( (0.5)(0.5)/20 )

Reject if x/20   < (-1.645) sqrt ( (0.5)(0.5)/20 ) + 0.5

Reject if x/20   < 0.316

Reject if x   < (0.316)(20) = 6.32

{0,1,2,3,4,5,6} is the region for the best level 0.05 test

c)

According to (a),  reject H0 if x <= 5

P( Type II error) = P( do not reject H0/ when Ha is true)

P( Type II error) = P( x > 5/ P=0.6)

x ---p(x)

6  0.004854  0.998388  

7  0.014563    

8  0.035497  

9  0.070995  

10  0.117142  

11  0.159738  

12  0.179706  

13  0.165882  

14  0.124412  

15  0.074647  

16  0.034991  

17  0.012350  

18  0.003087  

19  0.000487  

20  0.000037  

add: 0.9984 --  proba bility of a type II error

Assuming P=0.8

P( Type II error) = P( x > 5/ P=0.8)

6  0.000002  1.000000  

7  0.000013  

8  0.000087  

9  0.000462  

10  0.002031  

11  0.007387  

12  0.022161  

13  0.054550  

14  0.109100  

15  0.174560  

16  0.218199  

17  0.205364  

18  0.136909  

19  0.057646  

20  0.011529  

add: 1.0000  probability of a type II error

d)

P( x <= 13) =  

0  0.000001  

1  0.000019  

2  0.000181  

3  0.001087  

4  0.004621  

5  0.014786  

6  0.036964  

7  0.073929  

8  0.120134  

9  0.160179  

10  0.176197  

11  0.160179  

12  0.120134  

13  0.073929  

add: 0.9423 < 0.10 ,  H0 cannot be rejected

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3 years ago
Write one digit on each side of 23 to make a four digit multiple of 72. How many different solutions does this problem have?
user100 [1]

Answer:

This problem has two possible solutions.

2232

7236

Step-by-step explanation:

To be divisible by 72, a number has:

To be divisible by both 8 and 9 at the same time.

It is divisible by 8 if it's last two digits are divisible by 4.

It is divisible by 9 it the sum of it's digits is a number divisible by 9.

a23b

3b must be divisible by 4. In the thirties, the numbers that are divisible by 4 are 32 and 36. So b = 2 or b = 6.

a232

2 + 3 + 2 = 7

The higest possible value of a is 9 and the lowest is 1.

Between 8 and 16, only 9 is divisible by 9. So a = 2.

a = 2, b = 2 is one of the solutions.

a236

2 + 3 + 6 = 11

The higest possible value of a is 9 and the lowest is 1.

Between 12 and 20, only 18 is divisible by 9. So we need a = 7

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3 years ago
Decompose49.67 to to the nearest tenth
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49.7 because on work.com
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