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ICE Princess25 [194]
2 years ago
5

In ΔWXY, w = 880 cm, ∠X=33° and ∠Y=38°. Find the length of y, to the nearest centimeter.

Mathematics
1 answer:
vladimir2022 [97]2 years ago
3 0

Answer:

y\approx573cm

Step-by-step explanation:

First, take a look to the picture that I attached,  however please note the triangle is not drawn to scale, the figure is just to provide visual aid. As you can see the value of \angle W =109^{\circ} this is because of the sum of the interior angles in a triangle is always equal to 180°. So:

\angle W + \angle X + \angle Y =180\\\\\angle W = 180- \angle X -\angle Y\\\\\angle W =180-38-33\\\\\angle W=109

Now, we can use the law of sines, which states:

\frac{w}{sin(W)} =\frac{x}{sin(X)} =\frac{y}{sin(Y)}

Hence:

\frac{w}{sin(W)} =\frac{y}{sin(Y)}\\\\\frac{880}{sin(109)} =\frac{y}{sin(38)}\\\\Solving\hspace{3}for\hspace{3}y\\\\y=\frac{880*sin(38)}{sin(109)} \\\\y=572.9999518\approx 573 cm

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Answer:

-42

Step-by-step explanation:

negative * negative = positive

positive * negative = negative

negative * positive = negative

(-6)(-7)(-1) =

= 42(-1)

= -42

5 0
2 years ago
Scores on an exam are normally distributed with a mean of 76 and a standard deviation of 10. In a group of 230 tests, how many s
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The score of 96 is 2 standard deviations above the mean score. Using the empirical rule for a normal distribution, the probability of a score above 96 is 0.0235.
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2 years ago
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A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 mm and standard deviation 0.08
marta [7]

Answer:

(a) The proportion of the diameters are less than 25.0 mm is 0.1056.

(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

  ≈ 24.99

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.

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Answer:

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Then we plug in x to find numeric value.

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2 years ago
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