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8_murik_8 [283]
4 years ago
8

Find the value of x and pls explain!!❤❤❤

Mathematics
2 answers:
andreev551 [17]4 years ago
6 0
X=23.75
I know this because 3x-590x =180
4x 85=180
4x95
exis [7]4 years ago
4 0
3x - 5 + 90 + x = 180
4x + 85 = 180
4x = 95
x = 23.75
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There are 75 students enrolled in a camp. The day before the camp begins, 8% of the students cancel. How many students actually
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Step-by-step explanation:

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What is the mean, median, mode, and range of 8, 35, 10, 12, 14?
Ivan

answer:

mean: 15.8

median: 12

mode: no mode

range: 27

step-by-step explanation:

  • first know how to find each of the above and what they are
  • mean = (avg) add all the numbers, then divide by the number of numbers
  • median = the middle number from increasing order
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  • range = the difference between the largest and smallest number

8, 35, 10, 12, 14

<u>mean</u>

8 + 35 + 10 + 12 + 14 = 79

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79 / 5 = 15.8

<u>median</u>

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8, 10, 12, 14, 35

  • 12 is the median OR the middle number

<u>mode</u>

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3 years ago
What is 1.3333333 as a fraction ?
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2) Calculate the surface<br> area of the cylinder:<br> 5 cm
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Answer:

28π cm²

Step-by-step explanation:

The surface of a cylinder is given by the formula:

  • 2πrh + 2πr²

where r is the radius of the base, and h is the height.

Using the image, we can find that the radius of the cylinder would be 2 cm, and the height would be 5 cm.

Now, to find the surface area of the cylinder, we would have to plug in 2 for r (the radius of the base) and 5 for h (the height of the cylinder) into the formula for the surface area of a cylinder.

Now we plug in 2 for r and 5 for h for the equation for the surface area of a cylinder:

2π(2)(5) + 2π(2)² = 20π + 2π(4) = 20π + 8π = 28π

Remember, the radius and the height are given to us in cm, and w are solving for surface area, so the answer should be in cm² (square centimeters).

The surface area of the cylinder would be 28π cm².

28π cm² would be an exact answer. For the approximate answer, plug in 3.14 (most math classes use 3.14 to approximate π) for π.

I hope you find this helpful. :)

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