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nasty-shy [4]
1 year ago
5

Simplify 12+(38−18)2

Mathematics
1 answer:
Basile [38]1 year ago
7 0

Answer:

52

Step-by-step explanation:

12 + (38 - 18)2 = 12 + 20 × 2 = 52.

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Which of the following values of r will result in a true statement when substituted into the given equation?
Charra [1.4K]

Answer:

the right answer is -5 (B)

Step-by-step explanation:

-(-5) +5(-5+2)

5 + -15 = -10

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3 years ago
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The temperature in your town is 31 degrees Fahrenheit. The radio announcer says that the temperature will drop 15 degrees.What w
laiz [17]

Hi,

New temperature will be 31 F - 15 F = 16 F.

Green eyes.

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write the equation of the line (in slope-intercept form) that passes through the points (2,3 and (5,7)
AfilCa [17]

Answer:

The answer to your question is    y = 4/3x + 1/3

Step-by-step explanation:

Data

Point A = (2, 3)

Point B = (5, 7)

Process

1.- Calculate the slope

x1 = 2    y1 = 3

x2 = 5    y2 = 7

m = (y2 - y1)/(x2 - x1)

- Substitution

  m = (7 - 3)/(5 - 2)

- Slope

  m = 4/3

2.- Find the equation of the line

     y - y1 = m(x - x1)

     y - 3 = 4/3(x - 2)

     y - 3 = 4/3x - 8/3

     y = 4/3x - 8/3 + 3

     y = 4/3x - 8/3 + 9/3

     y = 4/3x + 1/3

8 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
What is answer to the questions
puteri [66]

Answer:

Meaning

Step-by-step explanation:

7 0
3 years ago
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