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lisov135 [29]
3 years ago
9

A student has scores of 85,83,98 and 77 on four quizzes. What must she score on the fifth quiz to have an average of at least 84

?
Mathematics
1 answer:
natulia [17]3 years ago
7 0
We need to find the solution to:
(85 + 83 + 98 + 77 + x) / 5 >= 84 where x is the score in the last test
This is the same as:
(343 + x) / 5 >= 84
Multiplying both sides by 5 we get
343 + x >= 420
Subtracting 343 from both sides we get
x >= 77

There for the student must get a score of at least 77 on the last tests.
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Step-by-step explanation:

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3 years ago
100 x 54 + 77 ÷ 2 = ? (RIGHT ANSWER GETS BRAINLIEST)
Nadusha1986 [10]

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5438.5

Step-by-step explanation:

To complete this equation, we must use the order of operations also known as PEMDAS.

For the case of this problem, we start by multiplying or dividing numbers before adding or subtracting numbers.

First let's start on the left:

100 x 54 + 77 ÷ 2

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5438.5

I hope this helps! Just remember to always add or subtract number last and start on the left, so that you don't get confused.

6 0
3 years ago
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A sphere is inscribed in a cube with a surface area of 216 square centimeters. What is the volume of the sphere
pentagon [3]

Answer:

V=298.5cm^3

Step-by-step explanation:

To find the volume of the sphere we need to know its radius.

And the radius can be found with the information we have about the surface area.

The formula for the surface area is as follows:

SA=4\pi r^2

from this formula we clear the radius r:

r^2=\frac{SA}{4\pi}\\ \\r=\sqrt{\frac{SA}{4\pi} }

and we substitute the known value of the surface area, and also the value of \pi:

r=\sqrt{\frac{216cm^2}{4(3.1416)} }\\ \\r=\sqrt{\frac{216cm^2}{12.5664} }\\ \\r=\sqrt{17.1887cm^2}\\ \\r=4.146cm

and now that we know the radius we find the volume:

V=\frac{4\pi r^3}{3} \\\\V=\frac{4(3.1416)(4.146cm)^3}{3}\\ \\V=\frac{895.521cm^3}{3}\\ \\V=298.5cm^3

the volume of the sphere is 298.5cm^3

7 0
2 years ago
Use the divergence theorem to find the outward flux of the vector field F(x,y,z)=2x2i+5y2j+3z2k across the boundary of the recta
aksik [14]

Answer:

The answer is "120".

Step-by-step explanation:

Given values:

F(x,y,z)=2x^2i+5y^2j+3z^2k \\

differentiate the above value:

div F =2 \frac{x^2i}{\partial x}+5 \frac{y^2j}{\partial y}+3 \frac{z^2k }{\partial z}  \\

        = 4x+10y+6z

\ flu \ of \ x = \int  \int div F dx

              = \int\limits^1_0 \int\limits^3_0 \int\limits^1_0 {(4x+10y+6z)} \, dx \, dy \, dz \\\\ = \int\limits^1_0 \int\limits^3_0 {(4xz+10yz+6z^2)}^{1}_{0} \, dx \, dy  \\\\ = \int\limits^1_0 \int\limits^3_0 {(4x+10y+6)} \, dx \, dy  \\\\ = \int\limits^1_0  {(4xy+10y^2+6y)}^3_{0} \, dx   \\\\ = \int\limits^1_0  {(12x+90+18)}\, dx   \\\\= {(12x^2+90x+18x)}^{1}_{0}   \\\\= {(12+90+18)}   \\\\=30+90\\\\= 120

6 0
3 years ago
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Natalija [7]
Side 1: 17 Cm.
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Hope This Helped.
8 0
2 years ago
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