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2 years ago

11

In a certain city, the hourly wage of workers on temporary employment contracts is normally distributed. The mean is $15 and the

standard deviation is $3. What percentage of temporary workers earn less than $12 per hour?

1 answer:

dmitriy555 [2]2 years ago

6 0

Answer:

15.87%

Step-by-step explanation:

We have been given that in a certain city, the hourly wage of workers on temporary employment contracts is normally distributed. The mean is $15 and the standard deviation is $3.

Let us find the z-score of 12, using z-score formula.

$z=\frac{x-\mu}{\sigma}$ , where,

$z$ = z-score,

$x$ = Random sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

Upon substituting our given values in z-score formula we will get,

$z=\frac{12-15}{3}$

$z=\frac{-3}{3}$

$z=-1$

Let us find P(z<-1) using normal distribution table.

$P(z$

Therefore, probability of a temporary worker earns less than $12 per hour is 0.15866. Let us convert our given probability in percentage by multiplying by 100.

$\text{The percentage of temporary workers earn less than \$12 per hour}=0.15866\times 100$

$\text{The percentage of temporary workers earn less than \$12 per hour}=15.866\approx 15.87$

Therefore, 15.87% of temporary workers earn less than $12 per hour.

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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tangare [24]

<u>Answer:</u>

The correct answer option is B. 21.

<u>Step-by-step explanation:</u>

We are given a figure where the line KL is intersecting another line MN such that angle from K to N is equal to the angle opposing it which is the angle measuring $(5x)$ .

Measure of angle K - N = 90° + 15° = 105°

Since angle K - N = angle M - L, we can write it as:

105° = 5x

x = 105/5

x = 21

Therefore, the correct answer option is B. 21.

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