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Aleks04 [339]
3 years ago
7

Which values are solutions to the inequality below?

Mathematics
2 answers:
goldenfox [79]3 years ago
6 0

Answer:

80, 75

Step-by-step explanation:

\sqrt{x} < 9

x < 9²

x < 81

Thus, 8 is lesser than 81. 80 and 75 are the only numbers lesser than 81 so they are the answer.

lozanna [386]3 years ago
3 0

Answer:

is 5 i hope it helps

Step-by-step explanation:

okay remember that this < stand for less numbers

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AlexFokin [52]
2(a+3)+3(2a-1)
=2a+6+6a-3
=8a+3

7 0
3 years ago
The difference between two integers is at most 16. the smaller integer is 12. what is the larger integer
emmainna [20.7K]
<span>The difference between two integers is at most 16
x1 - big integer
x2- small integer
x_1 -x_2  \leq 16

</span><span>the smaller integer is 12
so
x2 = 12

substitution
</span>
<span>x_1 -12  \leq 16

solve for x1

however, note the fact larger
thus meaning
x1 > x2 applies to this problem
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3 0
3 years ago
*I thought of a number, added 18, then multiplied the result by 8. I got 16. What was my number?
Arte-miy333 [17]

Answer:

-16

Step-by-step explanation:

Represent the number by n.

Then:

8(n + 18) = 16

Performing the indicated multiplication, we get:

8n + 144 = 16

Subtracting 144 from both sides results in 8n = -128.

Dividing both sides by 8:  n = -128/8  =  -16

The number was -16.

4 0
4 years ago
which statement about the points (2, 5) and (-2, 5) is true plot the points on a coordinate plane to help you answer the questio
iVinArrow [24]

Answer:

A, (2,5) and (-2,5) are reflections of each other on the x axis

5 0
3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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