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sveticcg [70]
3 years ago
8

Solving for 10x +14 = -2x+38

Mathematics
2 answers:
dalvyx [7]3 years ago
6 0

10x + 14 = -2x + 38

Subtract 14 from both sides

10x = -2x + 24

Add 2x to both sides

12x = 24

Divide by 12 on both sides

x = 2

ANSWER:

x = 2

natta225 [31]3 years ago
4 0

Answer:

2

Step-by-step explanation:

10x+2x=12x

38-14= 24

12x=24

divide

x=2

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Step-by-step explanation:

since both objects and sides are similar, you can set each side equal, then you cross multiply, and do subtraction, then find x.

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3 years ago
The volume of a cuboid is 176cm
IRISSAK [1]

Answer:

.08cm or 0.08cm

Step-by-step explanation:

First of all, we can see that 20mm is in millimeters, not centimeters.

If we convert the ones in centimeters to millimeters, it will result in decimals, which I want to avoid, but you could also do it that way.

There are 10 mm in 1 cm, so I multiply all of the values in cm by 10.

So now, we have that the volume is 1760mm, the length is 110mm, and the width is 20mm.

The formula for the volume of a cuboid is: w*l*h = volume

We have the width, length, and volume, so let's plug it in.

20*110*h=1760

To solve for h:

1. Multiply 20 and 100

2200*h=1760

2. Divide both sides by 2200

h = 1760/2200

3. Divide right side

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6 0
3 years ago
A new battery's voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries wil
klio [65]

Answer:

The probability that you test exactly 4 batteries is 0.0243.

Step-by-step explanation:

We are given that a new battery's voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found.

Suppose that 90% of all batteries have acceptable voltages.

Let the probability that batteries have acceptable voltages = P(A) = 0.90

So, the probability that batteries have unacceptable voltages = P(U) = 1 - P(A) = 1 - 0.90 = 0.10

Now, the probability that you test exactly 4 batteries is given by the three cases. Firstly, note that batteries will be tested until two acceptable ones have been found.

So, the cases are = P(AUUA) + P(UAUA) + P(UUAA)

This means that we have tested 4 batteries until we get two acceptable batteries.

So, required probability = (0.90 \times 0.10 \times 0.10 \times 0.90) + (0.10 \times 0.90 \times 0.10 \times 0.90) + (0.10 \times 0.10 \times 0.90 \times 0.90)

     =  0.0081 + 0.0081 + 0.0081 = <u>0.0243</u>

<u></u>

Hence, the probability that you test exactly 4 batteries is 0.0243.

3 0
3 years ago
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