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AVprozaik [17]
2 years ago
5

Help please i need help with my homework​

Mathematics
1 answer:
Sedbober [7]2 years ago
8 0

<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3

3 ( x ) + 55 = 85

Answer To Solving Y:

Let's solve your equation step-by-step.

3x + 55 = 85

Step 1: Subtract 55 from both sides.

3x + 55 − 55 = 85 − 55

3x = 30

Step 2: Divide both sides by 3.

3x / 3 = 30 / 3

x = 10

<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3

Answer To The Angle:

( 3 ) ( 10 ) + 55 = 85

<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3

2 ( y ) - 5 = 95

Answer:

Let's solve your equation step-by-step.

2y − 5 = 95

Step 1: Add 5 to both sides.

2y − 5 + 5 = 95 + 5

2y = 100

Step 2: Divide both sides by 2.

2y / 2 = 100 / 2

y = 50

<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3

Answer To The Angle:

( 2 ) ( 50 ) − 5 = 95

<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3

Please don't come at me if I am wrong.....:\

<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3<3

But, the anwer should be 180 degrees

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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
The price of an item has dropped to
Deffense [45]
60% decrease.

$105-$42 = $63
$63/$105 = 0.6
0.6 x 100 = 60
8 0
3 years ago
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