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Molodets [167]
2 years ago
10

What is the answer of 156.256*458796.53

Mathematics
1 answer:
dybincka [34]2 years ago
5 0

Answer:

71689710.59168

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Two sections of a class took the same quiz. Section A had 15 students who had a mean score of 80, and Section B had 20 students
Ksju [112]

Answer:

86

Step-by-step explanation:

In this question, we are asked to calculate the average mean score for a group of students split into two different emerging groups.

We calculate the total score for each of the groups. This is done by multiplying the average score by the number of students.

Total score of section A is 15 * 80 = 1,200

For section B, total score is 20 * 90 = 1,800

Overall score is thus 1200 + 1800 = 3000

We add the scores together and divide by the total number of students in both sections

Average score is thus 3000/35 = 85.71 which is approximately 86

3 0
3 years ago
4. The figure represents a ramp. Find the height of the ramp.
IgorLugansk [536]
Notice that the triangle is a right triangle, because of the right angle. By the Pythagorean Theorem, a^2+b^2=c^2. Then 15^2+b^2=17^2\implies 225+b^2=289\implies b^2=64\implies b=8, d. 8 m
3 0
3 years ago
Read 2 more answers
What is the square root of 99 simplified​
kolezko [41]

Answer:

√ 99

Rewrite  99  as  

3 2 ⋅ 11 .

Tap for more steps...

√ 3 2 . 11

Pull terms out from under the radical.

3 √ 11

The result can be shown in multiple forms.

Exact Form:

3 √ 11

Decimal Form:

9.94987437

…

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
There are 9 classes of 25 students each, 4 teachers, and two times as many chaperones as teachers.
densk [106]

5 buses is the answer pls mark me brainliest

8 0
3 years ago
Aidan has 20 ft of fence with which to build a rectangular dog run. What is the maximum area he can enclose? Enter your answer i
irina1246 [14]
<h2>Maximum area is 25 m²</h2>

Explanation:

Let L be the length and W be the width.

Aidan has 20 ft of fence with which to build a rectangular dog run.

         Fencing = 2L + 2W  = 20 ft

                     L + W = 10

                      W = 10 - L

We need to find what is the largest area that can be enclosed.

     Area = Length x Width

      A = LW

       A = L x (10-L) = 10 L - L²

For maximum area differential is zero

So we have

      dA = 0

      10 - 2 L = 0

        L = 5 m

      W = 10 - 5 = 5 m

Area = 5 x 5 = 25 m²

Maximum area is 25 m²

7 0
3 years ago
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