To teach computer programming to employees, many firms use on the job training. A human resources administrator wishes to review the performance of trainees on the final test of the training. The mean of the test scores is 72 with a standard deviation of 5. The distribution of test scores is approximately normal. Find the z-score for a trainee, given a score of 82.

Answer:

c. 1.29

Step-by-step explanation:

Z-score:

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 p-value, we get the probability that the value of the measure is greater than X.

Andy 70578 60000 8200

This means that [tex]X = 70578, \mu = 60000, \sigma = 8200[/tex]

What is the z-score of Andy's salary?

This is Z, so:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70578 - 60000}{8200}[/tex]

[tex]Z = 1.29[/tex]

So the correct answer is given by option c.

Answer:

ಠ_ಠ

Step-by-step explanation:

Answer:

90

Step-by-step explanation:

360 ÷ 4 = 90

Answer:

B. <B is congruent to the <Z

OPTION B is the correct answer

Answer:

5

Step-by-step explanation:

took the test

The coordinates of the point that divides the line segment from J to K into a ratio of 2:3 are P(-5,7), and the y-coordinate is 7, the correct option is D.

Ratio is described as the comparison of two quantities to determine how many times one obtains the other. The proportion can be expressed as a fraction or as a sign: between two integers.

We are given that;

Ratio= 2:3

Now,

Let the point we are looking for be denoted as P(x,y), and let the ratio be 2:3, which means that the distance from J to P is 2/5 times the distance from J to K.

Using the distance formula, we can find the distance between J and K as:

d = sqrt((x2-x1)^2 + (y2-y1)^2)

= sqrt((-8 - (-3))^2 + (11 - 1)^2)

= sqrt(25 + 100)

= sqrt(125)

The distance from J to P is 2/5 of the total distance, which is:

(2/5)d = (2/5)sqrt(125) = 2sqrt(5)

Using the ratio formula, we get:

x = (3* (-3) + 2 * (-8)) / (3+2) = -5

y = (31 + 211) / (3+2) = 7

Therefore, by the given ratio answer will be 7.

Learn more about the ratio here:

brainly.com/question/13419413

#SPJ7

Answer: 10

Step-by-step explanation:

4 x 2 x 5 = 40

9 + 2 + 7 + 8 + 4 = 30

40 - 30 = 10

= 10

Answer:

This is a multiplication of functions g and f, and these functions have no restrictions(such as a even root or a fraction), and thus [tex](g \mult f)(0) = g(0)f(0) = -4(1) = -4[/tex]

Step-by-step explanation:

We are given the following functions:

[tex]f(x) = 1[/tex]

[tex]g(x) = x - 4[/tex]

Can you evaluate (g•f)(0)?

This is a multiplication of functions g and f, and these functions have no restrictions(such as a even root or a fraction), and thus [tex](g \mult f)(0) = g(0)f(0) = -4(1) = -4[/tex]

Answer:

To evaluate the composition, you need to find the value of function f first. But, f(0) is 1 over 0, and division by 0 is undefined. Therefore, you cannot find the value of the composition.

You must evaluate the function f first.

Division by 0 is undefined.

The composition cannot be evaluated.

Answer:

23

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

divide the area in 2 regions

4 x 2 = 8 (area of one region)

4 x 4 = 16 (area of second region)

8 + 16 = 24 (sum of areas of the two regions)

Answer:

wheres the underline pls let me know what is underlined ill answer it on comment

Answer:

D

Step-by-step explanation:

[tex]y - \frac{3}{3} + 12[/tex]

[tex]y - 1 + 12[/tex]

[tex]y + 11[/tex]

Hey there!

We are given two functions - one is Exponential while the another one is Linear.

[tex] \large{ \begin{cases} f(x) = {4}^{x} - 8 \\ g(x) = 5x + 6 \end{cases}}[/tex]

1. Operation of Function

[tex] \large{(f + g)(x) = f(x) + g(x)}[/tex]

2. Substitution

[tex] \large{(f + g)(x) = ( {4}^{x} - 8) + (5x + 6)}[/tex]

3. Evaluate/Simplify

[tex] \large{(f + g)(x) = {4}^{x} - 8 + 5x + 6} \\ \large{(f + g)(x) = {4}^{x} + 5x - 8 + 6} \\ \large{(f + g)(x) = {4}^{x} + 5x - 2}[/tex]

4. Final Answer

Answer:

The right solution is "0.5".

Step-by-step explanation:

According to the question,

P(pay with the sponsored credit card | order exceeds $110)

= [tex]\frac{P(Pay \ with \ the \ sponsored \ credit\ card\ and\ order\ exceeds\ 110)}{P(order \ exceeds \ 110)}[/tex]

= [tex]\frac{P(A \ and \ B)}{P(A)}[/tex]

By putting the values, we get

= [tex]\frac{0.17}{0.34}[/tex]

= [tex]0.5[/tex]

Thus, the above is the right solution.

Step-by-step explanation:

2tan 30° / 1 + tan² 30° =

2(⅓√3) /1 + (⅓√3)² =

⅔√3 / 1+ ⅓ =

⅔√3 / 4/3 =

2/4 √3 =

½√3 = sin 60° (proven)

Answer:

13,125 ducks

Step-by-step explanation:

The ratio of ducks:geese on the first day was:

175:63

On the last day (end of migration), he counted 4,725 geese.

To find the number of ducks using the same ratio, we are first going to divide 4,725 by 63 to find what number all the ducks and geese multiplied by:

4,725/63 = 75

The geese multiplied by 75. This means the ducks also multiplied by 75:

175*75 = 13,125

Barnaby counted 13,125 ducks.

Hope it helps (●'◡'●)

Answer:

[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]

[tex]P(x < 70) = 0.225[/tex]

Step-by-step explanation:

Given

[tex]Lower = 30[/tex]

[tex]Width = 10[/tex]

Solving (a): The relative frequency table

First, we construct the frequency table using the given parameters.

[tex]\begin{array}{cc}{Class} & {Frequency} &{30-39} & {1} & {40-49} & {1} & {50 - 59} & {2} & {60 - 69} & {5} & {70 - 79} & {13} & {80 - 89} & {10} & {90 - 99} & {8} & {Total} & {40}\ \end{array}[/tex]

The relative frequency (RF) is calculated as:

[tex]RF = \frac{Frequency}{Total}[/tex]

Using the above formula to calculate the relative frequency, the relative frequency table is:

[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]

Solving (b): [tex]P(x < 70)[/tex]

To do this, we add up the relative frequencies of classes less than 70.

i.e.

[tex]P(x < 70) = [30 - 39] + [40 - 49] + [50 - 59] + [60 - 69][/tex]

So, we have:

[tex]P(x < 70) = 0.025 + 0.025 + 0.050 + 0.125[/tex]

[tex]P(x < 70) = 0.225[/tex]

Answer:

Following are the response to the given points:

Step-by-step explanation:

For question 5.11:

For point a:

For all the particular circumstances, it was not an appropriate sampling strategy as each normal distribution acquired is at a minimum of 30(5) = 150 or 2.5 hours for a time. Its point is not absolutely fair if it exhibits any spike change for roughly 10 minutes.

For point b:

The problem would be that the process can transition to an in the state in less than half an hour and return to in the state. Thus, each subgroup is a biased selection of the whole element created over the last [tex]2 \frac{1}{2}[/tex] hours. Another sampling approach is a group.

For question 5.12:

This production method creates 500 pieces each day. A sampling section is selected every half an hour, and the average of five dimensions can be seen in a [tex]\bar{x}[/tex]line graph when 5 parts were achieved.

This is not an appropriate sampling method if the assigned reason leads to a sluggish, prolonged uplift. The difficulty would be that gradual or longer upward drift in the procedure takes or less half an hour then returns to a controlled state. Suppose that a shift of both the detectable size will last hours [tex]2 \frac{1}{2}[/tex] . An alternative type of analysis should be a random sample of five consecutive pieces created every [tex]2 \frac{1}{2}[/tex] hour.

Answer:

3

Step-by-step explanation:

make a column of x ,f, fx

then write income in x and no.of workers in f

andthen multiply both just like 100*3 ,100*2, 300*p, 400*2,500*1 write its answer fx

add the all fx and use this formula

mean =fx /n

260=adding total of fx divide by 5

Repeat same formula in no 2

Answer:

1.

a. 2

b. 0.1353 probability that exactly 0 customers will arrive during a five-minute period, 0.2707 that exactly 1 customer will arrive, 0.2707 that exactly 2 customers will arrive and 0.1805 that exactly 3 customers will arrive.

c. 0.1428 = 14.28% probability that delays will occur.

2.

a. 0.4512 = 45.12% probability that the service time is one minute or less.

b. 0.6988 = 69.88% probability that the service time is two minutes or less.

c. 0.3012 = 30.12% probability that the service time is more than two minutes.

Step-by-step explanation:

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

[tex]f(x) = \mu e^{-\mu x}[/tex]

In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.

The probability that x is lower or equal to a is given by:

[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]

Which has the following solution:

[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]

The probability of finding a value higher than x is:

[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]

Question 1:

a. What is the mean or expected number of customers that will arrive in a five-minute period?

0.4 customers per minute, so for 5 minutes:

[tex]\mu = 0.4*5 = 2[/tex]

So 2 is the answer.

Question b:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353[/tex]

[tex]P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707[/tex]

[tex]P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707[/tex]

[tex]P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1805[/tex]

0.1353 probability that exactly 0 customers will arrive during a five-minute period, 0.2707 that exactly 1 customer will arrive, 0.2707 that exactly 2 customers will arrive and 0.1805 that exactly 3 customers will arrive.

Question c:

This is:

[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]

In which:

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

The values we have in item b, so:

[tex]P(X \leq 3) = 0.1353 + 0.2707 + 0.2707 + 0.1805 = 0.8572[/tex]

[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.8572 = 0.1428[/tex]

0.1428 = 14.28% probability that delays will occur.

Question 2:

[tex]\mu = 0.6[/tex]

a. What is the probability that the service time is one minute or less?

[tex]P(X \leq 1) = 1 - e^{-0.6} = 0.4512[/tex]

0.4512 = 45.12% probability that the service time is one minute or less.

b. What is the probability that the service time is two minutes or less?

[tex]P(X \leq 2) = 1 - e^{-0.6(2)} = 1 - e^{-1.2} = 0.6988[/tex]

0.6988 = 69.88% probability that the service time is two minutes or less.

c. What is the probability that the service time is more than two minutes?

[tex]P(X > 2) = e^{-1.2} = 0.3012[/tex]

0.3012 = 30.12% probability that the service time is more than two minutes.

Answer:

9

Step-by-step explanation:

Answer:

huh ano yan huhu paki ayos ng sagot

Step-by-step explanation:

hahahhaa

Answer:

Length = 14 m, Width = 7 m

Step-by-step explanation:

Let the length is l and width is b.

Width, b = l-7

Area of the rectangle, A = 98 m²

We know that, the area of a rectangle is as follow :

[tex]A=lb[/tex]

So,

[tex]98=l(l-7)\\\\98=l^2-7l\\\\l^2-7l-98=0\\\\l^2+7l-14l-98=0\\\\l(l+7)-14(l+7)=0\\\\l=14,-7[/tex]

Length can't be negative. So,

Width, b = 14-7 = 7 m

So, the dimensions of the rectangle are 14 m and 7 m respectively.

Answer:

Step-by-step explanation:

Combination of 3 teachers out of 12:

Combination of 4 students out of 43:

Total combinations:

Step-by-step explanation:

here is the answer to your question

Answer:

using 2 below points to draw:

(0, 7)

(3.5, 6)

Step-by-step explanation:

using

Answer:

5.5inches

Step-by-step explanation:

1331/8=166.375

then length of a side is = cubic root of 166.375

=³√166.375

5.5