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wolverine [178]
3 years ago
5

What is the value of x in the following equation? 5x + 12 = 6(x + 1) A. 6 B. 5 C. 24 D. 7

Mathematics
1 answer:
Maurinko [17]3 years ago
4 0

Answer:

A: 6

Step-by-step explanation:

Our equation is 5x+12=6(x+1)

Our possible numbers are 6, 5, 24, or 7.

To figure this out you start putting the possible numbers in place of x so we'll start off using 6:

5*6+12=6(6+1).

5*6=30. 30+12=42. 6+1=7, 6*7=42.

This equals the same on both sides.

Next we'll go to 5:

5*5+12=6(5+1)

5*5=25, 25+12=37. 5+1=6, 6*6=36.

This doesn't equal the same amount on both sides.

Next is 24:

5*24+12=6(24+1)

5*24=120, 120+12=132. 24+1=25, 6*25=150

This doesn't equal the same amount.

Next is 7:

5*7+12=6(7+1)

5*7=35, 35+12=47. 7+1=8, 6*8=48.

This doesn't equal.

The only number that equals the same amount on both sides is A:6

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Leila says that 75% of a number will always be greater than 50% of any other number. Complete one inequality to support Leilas c
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If a number is positive, Leila's theory that 75% of a number will always be greater than 50% of another number is <em>true</em>;<em> </em>however, if both numbers are negative, or if the number of which she finds 50% is much greater than the number of which she finds 75%, Leila's theory could be incorrect.

This inequality shows that Leila is correct: 100(0.75) \geq 50(0.50) (which simplifies to 75 \geq 25)

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Hope this helps!
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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
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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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