Solve the system of equations 15x -2y = 4 20x

The U.S. Bureau of Labor Statistics released hourly wage figures for various countries for workers in the manufacturing sector.

elena-14-01-66 [18.8K]

Answer:

(a) The probability that the sample average will be between $30.00 and $31.00 is 0.5539.

(b) The probability that the sample average will exceed $21.00 is 0.12924.

(c) The probability that the sample average will be less than $22.80 is 0.04006.

Step-by-step explanation:

We are given that the hourly wage was $30.67 for Switzerland, $20.20 for Japan, and $23.82 for the U.S.

Assume that in all three countries, the standard deviation of hourly labor rates is $4.00.

(a) Suppose 40 manufacturing workers are selected randomly from across Switzerland.

Let $\bar X$ = sample average wage

The z score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean wage for Switzerland = $30.67

$\sigma$ = standard deviation = $4.00

n = sample of workers selected from across Switzerland = 40

Now, the probability that the sample average will be between $30.00 and $31.00 is given by = P($30.00 < $\bar X$ < $31.00)

P($30.00 < $\bar X$ < $31.00) = P( $\bar X$ < $31.00) - P( $\bar X$ $\leq$ $30.00)

P( $\bar X$ < $31) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{31-30.67}{\frac{4}{\sqrt{40} } }$ ) = P(Z < 0.52) = 0.69847

P( $\bar X$ $\leq$ $30) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{30-30.67}{\frac{4}{\sqrt{40} } }$ ) = P(Z $\leq$ -1.06) = 1 - P(Z < 1.06)

= 1 - 0.85543 = 0.14457

The above probability is calculated by looking at the value of x = 0.52 and x = 1.06 in the z table which has an area of 0.69847 and 0.85543 respectively.

Therefore, P($30.00 < $\bar X$ < $31.00) = 0.69847 - 0.14457 = 0.5539

(b) Suppose 32 manufacturing workers are selected randomly from across Japan.

Let $\bar X$ = sample average wage

The z score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean wage for Japan = $20.20

$\sigma$ = standard deviation = $4.00

n = sample of workers selected from across Japan = 32

Now, the probability that the sample average will exceed $21.00 is given by = P( $\bar X$ > $21.00)

P( $\bar X$ > $21) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{21-20.20}{\frac{4}{\sqrt{32} } }$ ) = P(Z > 1.13) = 1 - P(Z < 1.13)

= 1 - 0.87076 = 0.12924

The above probability is calculated by looking at the value of x = 1.13 in the z table which has an area of 0.87076.

(c) Suppose 47 manufacturing workers are selected randomly from across United States.

Let $\bar X$ = sample average wage

The z score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean wage for United States = $23.82

$\sigma$ = standard deviation = $4.00

n = sample of workers selected from across United States = 47

Now, the probability that the sample average will be less than $22.80 is given by = P( $\bar X$ < $22.80)

P( $\bar X$ < $22.80) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{22.80-23.82}{\frac{4}{\sqrt{47} } }$ ) = P(Z < -1.75) = 1 - P(Z $\leq$ 1.75)

= 1 - 0.95994 = 0.04006

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

3 0

3 years ago

Which graph shows the following system of equations and its solutions? -0.1x-0.3y=1.2 and 0.2x-0.5y=2

Vera_Pavlovna [14]

Answer:

Option b is the correct answer as both the equations are true for given solution.

Step-by-step explanation:

Given equations are:

-0.1x-0.3y=1.2

0.2x-0.5y=2

We can observe each graph and find the point that is the solution and put the point in the equations to know if that point is the solution

For option A:

(0,4)

Putting x=0 and y = 4 in both equations

$-0.1(0)-0.3(4)=1.2\\0-1.2 =1.2\\-1.2 \neq 1.2\\0.2(0)-0.5(4)=2\\0-2 = 2\\-2 \neq 2$

This is not the correct answer as both equations are not true with this solution

For Option B:

(0,-4)

Putting x = 0 and y = -4 in both equations

$-0.1(0)-0.3(-4)=1.2\\0+1.2 = 1.2\\1.2 = 1.2\\0.2x-0.5y=2\\0.2(0)-0.5(-4) = 2\\2 = 2$

Both equations are true for (0,-4) hence it is the solution of the system.

For Option C:

(4,0)

$-0.1(4)-0.3(0)=1.2\\-0.4-0 = 1.2\\-0.4 \neq 1.2\\0.2x-0.5y=2\\0.2(4)-0.5(0) = 2\\0.8 \neq 2$

Not true for both equations

Hence,

Option b is the correct answer.

7 0

2 years ago

Please awnser quickly

lakkis [162]

Answer:

B': (-8,2)

C':(-8,9)

D':(-5,4)

Step-by-step explanation:

If im wrong pls tell me!

7 0

2 years ago

if an office of 36 typist uses 12 dozen envelopes per day how many dozen envelope would an office of 125 typists of use?

Pani-rosa [81]

Assuming the envelope consumption is proportional to the number of typists, you have ...
(12 dz)/36 = x/125
125*(12 dz)/36 = x = 41 2/3

An office of 125 typists would use 41 2/3 dozen envelopes per day.

6 0

3 years ago

Dose this table represent a function?

iogann1982 [59]

No it doesnt because it 8 is repeated and function has 1 outcome for each x

8 0

3 years ago

Solve the system of equations 15x -2y = 4 20x - 3y = 1