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2 years ago

Use the graph of the function f to determine

Mathematics

1 answer:

FromTheMoon [43]2 years ago

8 0

Should be B
The first 2 just look at what y value corresponds to x value in the limit, and for the last one it’s DNE because at -2 the limit approaches 1 and -1 which means it’s undefined and uncertain

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The average wage for an electrician in CA is $25.47/hour. That is 11% above the national average. What is the national average?

yan [13]

Answer:

$22.95/hour.

Step-by-step explanation:

It is given that the average wage for an electrician in CA is $25.47/hour. That is 11% above the national average.

Let as consider x be the national average.

11% above the national average is

$x+\dfrac{11}{100}x=x+0.11x=1.11x$

The average wage for an electrician in CA is 11% above the national average. So,

$1.11x=25.47$

Divide both sides by 1.11.

$x=\dfrac{25.47}{1.11}$

$x=22.945945...$

$x=22.95$

Therefore, the national average is $22.95/hour.

3 0

3 years ago

Through (4,-4) ; perpendicular to 6y=x-12

BARSIC [14]

y = mx + b

m = slope and b = y-intercept

We can arrange 6y = x - 12 in the form of y = mx + b

6y = x - 12

y = 1/6(x) - 2

Slope of y = 1/6(x) - 2 is 1/6. Taking the negative reciprocal of the slope we get the slope for the perpendicular line.

Negative reciprocal of 1/6 is -6.

The equation for the perpendicular line is

y = -6x + b

To find b we can plug in the x and y values of (4,-4) into it since it passes through those coordinates

-4 = -6(4) + b

b = -4 + 6(4)

b = -4 + 24

b = 20

So the equation for the perpendicular line is y = -6x + 20

3 0

3 years ago

A purchaser of electric relays buys from two suppliers, A and B. Supplier A supplies two of every three relays used by the compa

inessss [21]

Answer:

The probability that at most 38 of the 60 relays are from supplier A is P(X≤38)=0.3409.

Step-by-step explanation:

We can model this question with a binomial distribution random variable.

The sample size is n=60.

The probability that the relay come from supplier A is p=2/3 for any relay.

If we use a normal aproximation, we have the mean and standard deviation:

$\mu=np=60*(2/3)=40\\\\\sigma=\sqrt{npq}=\sqrt{60*(2/3)*(1/3)} =3.65$

The probability that at most 38 of the 60 relays are from supplier A is P(X≤38)=0.3409:

$P(X\leq38)=P(X$

5 0

2 years ago

What is the first step in evaluating the expression shown below?

Olegator [25]

Answer:

Step-by-step explanation:

because it is is parenthesis

8 0

3 years ago

4. The distribution of blood cholesterol level in the population of young men aged 20 to 34 years is close to Normal, with mean

Pie

Answer:

a) 38.59% probability that a young man (aged 20 to 34) has a cholesterol level greater than 200 milligrams per deciliter.

b) By the Central Limit Theorem, the mean of the distribution of the sample mean would be 188 milligrams per deciliter

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 188, \sigma = 41$

a. Find the probability that a young man (aged 20 to 34) has a cholesterol level greater than 200 milligrams per deciliter.

This is 1 subtracted by the pvalue of Z when X = 200. So

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

$Z = \frac{200 - 188}{41}$

$Z = 0.29$

$Z = 0.29$ has a pvalue of 0.6141

1 - 0.6141 = 0.3859

38.59% probability that a young man (aged 20 to 34) has a cholesterol level greater than 200 milligrams per deciliter.

b. Suppose you measure the cholesterol level of 100 young men chosen at random and calculate the sample mean. If you did this many times, i. what would be the mean of the distribution of the sample mean

By the Central Limit Theorem, the mean of the distribution of the sample mean would be 188 milligrams per deciliter

5 0

2 years ago