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3 years ago

If line m is perpendicular to line n, and the slope of line m is -5/3, what is the slope of line n? Write in simplest form.

Mathematics

1 answer:

frez [133]3 years ago

7 0

Answer:

Step-by-step explanation:

Slope of perpendicular line = $\frac{-1}{m}$

= -1 ÷ $\frac{-5}{3}$

$= -1 *\dfrac{-3}{5}\\\\=\dfrac{3}{5}$

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Find the value of x in the triangle shown below.<br> 13<br> 5<br> х

katen-ka-za [31]

Answer:

x=12

Step-by-step explanation:

This is a right triangle, so the Pythagorean theorem can be used. The Pythagorean theorem states that $C^2=a^2+b^2$ where C is the hypotenuse and a and b are legs. In this question, we are given a and C. So, plugin 13 for C and 5 for a and solve for b. Once you rearrange the equation you get, $b=\sqrt{13^2-5^2}$ . Next, solve what is under the root to get, $b=\sqrt{144}$ . Finally, square root 144 for the final answer b=12.

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3 years ago

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Find the simple interest on #1500 in 5 years at 4% per annum

ioda

Si is = PRT so your answer is 1500x5x4 which is = 30000

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4 years ago

What is the answer to this 28-4x-3+3x

nikitadnepr [17]

25-x is the answer :)

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3 years ago

Does the formula V = 4/3 pi r^3 represent a linear or nonlinear function? Explain.

Levart [38]

This is the volume of the sphere & is not at all a linear function.

A linear function of the form of y=kx (k is a given number/coefficient) where

y & x grow at the same rate. Then whatever the input (x) it will generate an output (y) always in the ration y/x, whereas v=4/3 πr³ (coefficient =k =4/3(π), if r increase by 2 units v will increase by 2³ =8 & for r = 3, v increase by 3³ =27.

In short v increases quicker than r. No it's not a linear function

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3 years ago

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production

Ghella [55]

Answer:

a) 0.3174 = 31.74% probability of a defect. The number of defects for a 1,000-unit production run is 317.

b) 0.0026 = 0.26% probability of a defect. The expected number of defects for a 1,000-unit production run is 26.

c) Less variation means that the values are closer to the mean, and farther from the limits, which means that more pieces will be within specifications.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Assume a production process produces items with a mean weight of 10 ounces.

This means that $\mu = 10$ .

Question a:

Process standard deviation of 0.15 means that $\sigma = 0.15$

Calculate the probability of a defect.

Less than 9.85 or more than 10.15. Since they are the same distance from the mean, these probabilities is the same, which means that we find 1 and multiply the result by 2.

Probability of less than 9.85.

pvalue of Z when X = 9.85. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{9.85 - 10}{0.15}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

2*0.1587 = 0.3174

0.3174 = 31.74% probability of a defect.

Calculate the expected number of defects for a 1,000-unit production run.

Multiplication of 1000 by the probability of a defect.

1000*0.3174 = 317.4

Rounding to the nearest integer,

The number of defects for a 1,000-unit production run is 317.

Question b:

Now we have that $\sigma = 0.05$

Probability of a defect:

Same logic as question a.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{9.85 - 10}{0.05}$

$Z = -3$

$Z = -3$ has a pvalue of 0.0013

2*0.0013 = 0.0026

0.0026 = 0.26% probability of a defect.

Expected number of defects:

1000*0.0026 = 26

The expected number of defects for a 1,000-unit production run is 26.

(c) What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

Less variation means that the values are closer to the mean, and farther from the limits, which means that more pieces will be within specifications.

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3 years ago