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Amiraneli [1.4K]
2 years ago
9

Is x=12 a solution to the given inequality x > 5

Mathematics
1 answer:
Ludmilka [50]2 years ago
4 0

Answer: True

Step-by-step explanation:

Substitute 12 for x in the inequality

x > 5

12 > 5

12 is greater than 5 so x = 12 is a solution to x > 5

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9+15÷3(2)-6<br> Please help me
Firlakuza [10]

Answer:

13

Step-by-step explanation:

4 0
3 years ago
Help please !!! At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating
Radda [10]

Answer:

n = 12$    x = 8$   y = 10$     where n, x, and y are number of tickets

12 n + 8 x + 10 y = 3920     and n + x + y = 420

12n + 8 (x + y) + 2 y = 3920

12 n + 8 (5 n) + 2 y = 3920        since 5 (x + y) = n

52 n + 2 y = 3920   or  y = 1960 - 26 n

Also, n + x + y = 420   or n + 5 n = 420   since x + y = 5 n

n = 70    so 70 of the $12 were sold

And since y = 1960 - 26 n     we have y = 140 tickets

Now 12 * 70 + 8 x + 140 * 10 = 3920

This gives x = 210 tickets

Check:   210 + 140 + 70 = 420 tickets

Also, 12 * 70 + 210 * 8 + 140 * 10 = 3920

7 0
3 years ago
N=4, A=81<br> The roots are ? And ?
Neko [114]

Answer:±2 and±9

Step-by-step explanation:

±√(4) = ±2

±√(81) = ±9

7 0
3 years ago
NO LINKS PLEASE- OR I SWEAR-
lesantik [10]

Answer:

60

Step-by-step explanation:

To find the mean;

add up the data

(50+60+60+70) =240

Then divide by the number of data points

240/4 = 60

The mean is 60

5 0
3 years ago
Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial t
jekas [21]

Answer:

(12.1409, 14.0591

Step-by-step explanation:

Given that Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known.

Group   Group One     Group Two  

Mean 87.600 74.500

SD 1.000 1.500

SEM 0.316 0.433

N 10      12    

The mean of Group One minus Group Two equals 13.100

standard error of difference = 0.556

 90% confidence interval of this difference:  

(13.1-1.725*0.556,13.1+1.725*0.556)\\=(12.1409, 14.0591)

  t = 23.5520

 df = 20

4 0
3 years ago
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