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serg [7]
2 years ago
10

Evaluate. 2 • {[5 • (60 – 14 ÷ 7)] + 25} • 5

Mathematics
1 answer:
Mrac [35]2 years ago
6 0
The answer is 3150 :)
BODMAS
brackets, division, multiplication, addition, subtraction
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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 = <u>sample average wage</u>

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

<em>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.</em>

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

<u></u>

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

Let \bar X = <u>sample average wage</u>

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 = <u>0.12924</u>

<em />

<em>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.</em>

<em />

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

Let \bar X = <u>sample average wage</u>

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 = <u>0.04006</u>

<em />

<em>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.</em>

3 0
3 years ago
The expression 9+21 factored using GCF is?
Andreas93 [3]

The GCF is 3. 9 is 3 and 3 ; 21 is 7 and 3

7 0
3 years ago
Factorise: a squared +8a -20
Vaselesa [24]
Factor out the 4 in both equations
8a^2-20^2=(2^2 times a^2 times 2)-(2^2 times 5)
therefor it is also equal to
(2a)^2 times 2-(2^2 times 5)
we can force it into a difference of 2 perfect squares which is a^2-b^2=(a-b)(a+b)
(2a√2)^2-(2√5)^2=((2a√2)-(2√5))((2a√2)+(2√5))


8 0
3 years ago
Write the three numbers less then 40 that have 2 and 5 as factors
ozzi
No. which have factors 2 & 5 have 2 & 5 as their prime factor
therefore, No. are :
2*5 = 10
2*5*2 = 20
2*5*3 = 30
3 0
3 years ago
Kong took 15\%, percent fewer seconds than Nolan took to complete his multiplication timed test. Kong took 85 seconds. How many
andreev551 [17]

Answer:

Nolan took 97.75 seconds. you may need to round it to 98.

Step-by-step explanation:

multiply 15% by 85 which equals 12.75 then add to 85

6 0
3 years ago
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