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xz_007 [3.2K]
2 years ago
11

Karinas science test scores for this quarter are 84,86,90 and 68. What score does she need on her fifth science test to get a te

st average of at least 84
Mathematics
1 answer:
Ket [755]2 years ago
8 0

Answer:

Karina must score atleast 92 on the fifth test to get a average of atleast 84.

Step-by-step explanation:

We are given the following in the question:

84, 86, 90, 68

We want the average score to be atleast 84.

Let x be the score on fifth test.

Formula:

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Thus, we can write the equation:

\dfrac{84+86+90+68+x}{5}\geq 84\\\\\Rightarrow 84+86+90+68+x \geq 420\\\Rightarrow 328+x \geq 420\\\Rightarrow x \geq 92

Thus, Karina must score atleast 92 on the fifth test to get a average of atleast 84.

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How do you factor the common factor out of 30x^2-12x
Anika [276]
The answer would be 6x(5x-2).

Since both terms have x in them, x will be apart of your answer.

The GCF of 12 and 30 is 6.

So you can factor out 6x from both terms.

Hope this helps!
4 0
2 years ago
What points can be found on the line 2y=3x+4
Olin [163]

The answer is (2,5).

Check:

2y=3x+4

2(5)=3(2)+4

10=6+4

10=10 Good

The answer is (2,5)

3 0
3 years ago
Kaitlin is on a long car trip. Everytime she stops for gas she loses 15 minutes of travel time. If she has to stop fives times h
Nastasia [14]
She will be 75 minutes late because you take the 15 minutes per stop and multiply it by the five stops.
7 0
3 years ago
Read 2 more answers
A rectangular piece of metal is 25 in longer than it is wide. Squares with sides 5 in long are cut from the four corners and the
Licemer1 [7]

Answer:

The original length was 41 inches and the original width was 16 inches

Step-by-step explanation:

Let

x ----> the original length of the piece of​ metal

y ----> the original width of the piece of​ metal

we know that

When squares with sides 5 in long are cut from the four corners and the flaps are folded upward to form an open box

The dimensions of the box are

L=(x-10)\ in\\W=(y-10)\ in\\H=5\ in

The volume of the box is equal to

V=(x-10)(y-10)5

V=930\ in^3

so

930=(x-10)(y-10)5

simplify

186=(x-10)(y-10) -----> equation A

Remember that

The piece of metal is 25 in longer than it is wide

so

x=y+25 ----> equation B

substitute equation B in equation A

186=(y+25-10)(y-10)

solve for y

186=(y+15)(y-10)\\186=y^2-10y+15y-150\\y^2+5y-336=0

Solve the quadratic equation by graphing

using a graphing tool

The solution is y=16

see the attached figure

Find the value of x

x=16+25=41

therefore

The original length was 41 inches and the original width was 16 inches

3 0
3 years ago
Mr. Baldwin is packing the shipping box with ring boxes that are cubes with an edge length of 3/2 inches. If he completely fills
Alinara [238K]

Answer:

10 ring boxes

Step-by-step explanation:

First, we need to calculate the total surface area of each cube ring boxes

The surface area of each square boxes = 6L²

Given that L =1.5inches

Total surface area = 6(1.5)²

Total surface area = 6(2.25)

Total surface area = 13.5in²

<em>Since the question is incomplete. Let us assume the total surface area of the shipping box is 135in²</em>

<em></em>

Number of ring boxes he can ship = 135/13.5

Number of ring boxes he can ship = 10

Hence the number of ring boxes he can ship is 10 ring boxes

<em />

<u><em>NB: The total surface area of the shipping box was assumed</em></u>

<u><em></em></u>

6 0
3 years ago
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