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marissa [1.9K]
3 years ago
11

3x + 10 = 5(x – 17) how do i solve this? any easy examples?

Mathematics
1 answer:
Andreas93 [3]3 years ago
6 0

Answer:

x= 95/2  or  x=47.5  or   47 1/2

Step-by-step explanation:

3x + 10 = 5(x - 17)

First Step:

multiply and combine the right side (parentheses)

5 times x = 5x         5 times -17 = -85

3x + 10 = 5x -85

Second Step:

Add like terms ----> which is: 3x and 5x -----10 and -85

BUT since you only want one x on one side, you combine x first.

3x + 10 = 5x -85

-5x         -5x

-2x + 10 = -85

NOW

Third Step:     Add the integers---> 10 and -85

you want to get 10 on the same side as the -85, so you subtract it from BOTH sides.

which brings you:

-2x + 10 = -85

       -10    -10

<u>                        </u>

NOW that we've subtracted -10 from -85 it's technically just adding 85 and 10 then putting a negative sign on it.

so now we have:

-2x = -95

Step Four:

divide -2 from each side to get X alone

<u>-2x</u> = <u>-95</u>

-2      -2

ANSWER:

x=47.5

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(b) The 10th percentile of the diameters is 24.99 mm.

(c) The ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d) The proportion of the ball bearings meeting the specification is 0.8881.

Step-by-step explanation:

Let <em>X</em> = diameters of ball bearings.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 25.1 mm and standard deviation, <em>σ</em> = 0.08 mm.

To compute the probability of a Normally distributed random variable we need to first convert the raw scores to <em>z</em>-scores as follows:

<em>z</em> = (X - μ) ÷ σ

(a)

Compute the probability of <em>X</em> < 25.0 mm as follows:

P (X < 25.0) = P ((X - μ)/σ < (25.0-25.1)/0.08)

                    = P (Z < -1.25)

                    = 1 - P (Z < 1.25)

                    = 1 - 0.8944

                    = 0.1056

*Use a <em>z</em>-table for the probability.

Thus, the proportion of the diameters are less than 25.0 mm is 0.1056.

(b)

The 10th percentile implies that, P (X < x) = 0.10.

Compute the 10th percentile of the diameters as follows:

P (X < x) = 0.10

P ((X - μ)/σ < (x-25.1)/0.08) = 0.10

P (Z < z) = 0.10

<em>z</em> = -1.282

The value of <em>x</em> is:

z = (x - 25.1)/0.08

-1.282 = (x - 25.1)/0.08

x = 25.1 - (1.282 × 0.08)

  = 24.99744

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Thus, the 10th percentile of the diameters is 24.99 mm.

(c)

Compute the value of P (X < 25.2) as follows:

P (X < 25.2) = P ((X - μ)/σ < (25.2-25.1)/0.08)

                    = P (Z < 1.25)

                    = 0.8944

                    ≈ 0.84

*Use a <em>z</em>-table for the probability.

Thus, the ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d)

Compute the value of P (25.0 < X < 25.3) as follows:

P (25.0 < X < 25.3) = P ((25.0-25.1)/0.08 < (X - μ)/σ < (25.3-25.1)/0.08)

                    = P (-1.25 < Z < 2.50)

                    = P (Z < 2.50) - P (Z < -1.25)

                    = 0.99379 - 0.10565

                    = 0.88814

                    ≈ 0.8881

*Use a <em>z</em>-table for the probability.

Thus, the proportion of the ball bearings meeting the specification is 0.8881.

4 0
3 years ago
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