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## A random sample of 25 statistics examinations was taken. the average score in the sample was 76 with a standard deviation of 12. assuming the scores are normally distributed, the 99% confidence interval for the population average examination score is:

Since the sample size is less than 30, therefore we use
the t statistic.

Let us define the given variables:

N = sample size = 25

X = average score = 76

s = standard deviation = 12

99% Confidence interval

Degrees of freedom = n – 1 = 24

The formula for confidence interval is given as:

CI = X ± t * s / sqrt N

using the standard distribution table, the t value for DF
= 24 and 99% CI is:

t = 2.492

Therefore calculating the CI using the known values:

CI = 76 ± 2.492 * 12 / sqrt 25

CI = 76 ± 5.98

CI = 70.02, 81.98

Answer: The average score ranges from 70 to 82.

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## The grocery store sells kumquats for \$3.75 a pound and Asian pears for \$2.25 a pound. Write an equation in standard form for the weights of kumquats k and Asian pears p that a customer could buy with \$14.

an equation in standard form is Step-by-step explanation:

The Standard form of an equation is written as: here, k represents he weights of kumquats and p represents the weight of  Asian pears .

As per the statement:

The grocery store sells kumquats for \$3.75 a pound and Asian pears for \$2.25 a pound. We have to find the an equation in standard form

A customer could buy kumquats and Asian pears with \$ 14

then; Therefore, an equation in standard form is Categories

## Suppose that people’s heights (in centimeters) are normally distributed with a mean of 170 and a standard deviation of 5. We find the heights of 50 people.

We have
mean=mu=170
standard deviation=sigma=5

can now calculate the Zmin and Zmax using Z=(X-mean)/standard deviation
Zmin=(165-170)/5=-1
Zmax=(175-170)/5=+1
From normal probability tables,
P(z<Zmin)=P(z<-1)=0.15866
P(z<Zmax)=P(z<+1)=0.84134
P(165<x<175)=P(Zmin<z<Zmax)=0.84134-0.15866= 0.68269

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## Janitor supply produces an industrial cleaning powder that requires 40 grams of material at \$0.10 per gram and .25 direct labor hours at \$12.00 per hour. overhead is assigned at the rate of \$18 per direct labor hour. what is the total standard cost for one unit of product that would appear on a standard cost card? \$7.00. \$8.50. \$11.50. \$7.50. \$25.00.

Janitor supply produces an industrial cleaning powder that requires 40 grams of material at \$0.10 per gram and .25 direct labor hours at \$12.00 per hour. overhead is assigned at the rate of \$18 per direct labor hour. what is the total standard cost for one unit of product that would appear on a standard cost card? \$7.00. \$8.50. \$11.50. \$7.50. \$25.00.

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## A company is setting its direct materials and direct labor standards for its leading product. Direct material costs from the supplier areâ€‹ \$7 per squareâ€‹ foot, net of purchase discount.â€‹ Freight-in amounts toâ€‹ \$0.10 per square foot. Basic wages of the assembly line personnel areâ€‹ \$19 per hour. Payroll taxes are approximatelyâ€‹ 25% of wages. How much is the direct labor cost standard perâ€‹ hour

The formula for calculating the direct labor standard is:

Direct labor standard = Basic wages + Payroll tax

or

Direct labor standard = Basic wages + 25% of Basic wages

Substituting the given values into the equation:

Direct labor standard = \$19 + 0.25 * \$19

Direct labor standard = \$23.75

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## Sam is measuring the velocity of a car at different times. After two hours, the velocity of the car was 50km/h. After six hours, the velocity of the car is 54km/h. Part A: Write an equation in two variables in standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used. (5points) Part B: How can you graph the obtained in Part A for the first seven hours? (5points)

You are given the time and speed so from there you can get the points (2,50) and (6,54) once you got that you can see how the slope is 1 and
b=48
x-y=-48 (standard form)

part B=
1    49
2    50
3    51
4    52
5    53
6    54
7    55

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## Describe how you would determine whether two lines are parallel by looking at the equations of the lines in standard form.

By comparing the slopes of both lines.

Step-by-step explanation:

The general form of a straight line is y = mx + c where m is the slope of the line and c is the y-intercept of the line.

For e.g. Consider, the lines y = 2x – 3 and y = 2x – 7.

Now, by comparing these lines with the general form, we get that both of them have slope 2.

Also, by the graph below it can be seen that they both are parallel.

So, it is clear that ‘The slopes of two parallel lines are same’.

Hence, by looking at the standard form of the equation and comparing the slopes, we can determine whether the lines are parallel or not.

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## What price do farmers get for the peach crops? in the third week of June, a random sample of 40 farming regions gave a sample mean of \$6.88 per basket. assume that the standard deviation is known to be \$1.92 per basket. find a 90% confidence interval for the population mean price per basket that farmers in this region get for their peach crop

Given:
Sample size, n = 40
Sample mean, xb = \$6.88
Population std. deviation, σ = \$1.92 (known)
Confidence interval = 90%

Assume normal distribution for the population.
The confidence interval is
(xb + 1.645*(σ/√n), xb – 1.645*(σ/√n)
= (6.88 + (1.645*1.92)/√40, 6.88 – (1.645*1.92)/√40)
= (7.38, 6.38)

Answer: The 90% confidence interval is (7.38, 6.38)

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## Consider college officials in admissions registration, counseling, financial aid campus ministry, food services, and so on. How much money do these people make each year? Suppose you read in your local newspaper that 45 officials in student services earn an average of \$50,340 each year. Assume that the standard deviation is \$10,780 for salaries of college officials and student services. Find a 90% confidence interval for the population mean salaries of such personnel. Round your answer to the nearest dollar and don;t forget to use the \$ sign.

Since sample size is > 40, we use the z-score
in calculating for the confidence interval.

The formula is given as:

Confidence Interval = X ± z * σ / sqrt (n)

Where,

X = mean = \$50,340

z = z-score which is taken from standard distribution
tables at 90% confidence interval = 1.645

σ
= standard deviation = \$10,780

n = sample size = 45

Substituting to the equation:

Confidence Interval = 50,340 ± 1.645 * 10,780 / sqrt (45)

Confidence Interval = 50,340 ± 1,607

Confidence Interval = \$48,733 to \$51,947

Therefore the salary range of the personnel is \$48,733 to \$51,947.

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## 19. You roll two standard number cubes. What is the probability that the sum is odd, given than one of the number cubes shows a 1? Show your work.

there are 6 numbers per cube,

the chance of rolling one and getting the number 1 is a 1/6 probability

for the sum to be odd, you need to roll either a 2, 4 or 6 on the second cube so that gives you a 3/6 probability

so 1/6 x 3/6 = 3/36 which can be reduced to 1/12

so you have a 1/12 probability

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## In a standard dictionary, where can you find the key to pronunciation marks? A. At the bottom of each page B. Within the definitions C. In the front of the dictionary D. In an appendix

In a standard dictionary, where can you find the key to pronunciation marks? A. At the bottom of each page B. Within the definitions C. In the front of the dictionary D. In an appendix

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## Rewrite the slope-intercept form equation into standard form. y = –2x + 4 A. x + 2y = 4 B. 2x + y = 4 C. 4x + y = 2 D. 4x + 2y = 0 Please select the best answer from the choices provided

The answer is B because if you have y = -2x + 4, to get the -2x to the same side as y, you need to add, making it y + 2x = 4. It would also be postive no matter what because this formula needs to be positive. … and the standard form is Ax + By = C.. Making your final answer 2x + y = 4.

Hope This Helps!
Correct Me If I’m wrong!

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## The weights of certain machine components are normally distributed with a mean of 8.5 g and a standard deviation of 0.09 g. Find the two weights that separate the top 3% and the bottom 3%. These weights could serve as limits used to identify which components should be rejected. Round to the nearest hundredth of a gram.

To solve this problem,lets say that

X = the weight of the machine components.
X is normally distributed with mean=8.5 and sd=0.09

We need to find x1 and x2 such that
P(X<x1)=0.03 and P(X>x2)=0.03

Standardizing:

P( Z< (x1 – 8.5)/0.09 ) =0.03
P(Z > (x2 – 8.5)/0.09 ) =0.03.

From the Z standard table, we can see that approximately P
= 0.03 is achieved when Z equals to:

z = -1.88          and      z= 1.88

Therefore,

P(Z<-1.88)=0.03 and P(Z>1.88)=0.03

So,

(x1 – 8.5)/0.09 = -1.88 and
(x2 – 8.5)/0.09 =1.88

Solving for x1 and x2:

x1=-1.88(0.09) + 8.5   and
x2=1.88(0.09) + 8.5

Which yields:

x1 = 8.33 g

x2 = 8.67 g

Answer: The bottom 3 is separated by the weight
8.33 g and the top 3 by the weight 8.67 g.

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## Arrange 6x^2 – 2x^4 + 4 + 3x in standard form. I get the idea but I'm not sure… Options: –2x^4 + 6x^2 + 3x + 4 –2x^4 + 3x + 6x^2 + 4 6x^2 + 4 + 3x – 2x^4 4 + 3x + 6x^2 – 2x^4

Arrange 6x^2 – 2x^4 + 4 + 3x in standard form. I get the idea but I’m not sure… Options: –2x^4 + 6x^2 + 3x + 4 –2x^4 + 3x + 6x^2 + 4 6x^2 + 4 + 3x – 2x^4 4 + 3x + 6x^2 – 2x^4

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## Three cards are chosen from a standard deck of 52 playing cards with replacement. What is the probability every card will be an Ace? Can someone please help

Three cards are chosen from a standard deck of 52 playing cards with replacement. What is the probability every card will be an Ace? Can someone please help

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## Write he standard form of the equation; Through: (-1,5) and (1,2)

Standard form is ax+by=c where a,b, and c are integers and a is usually positive

so
first find slope
get into form y=mx+b where m is slope
so
slope between points (x1,y1) and (x2,y2) is
(y2-y1)/(x2-x1)
points (-1,5) and (1,2)
slope is (2-5)/(1-(-1))=-3/(1+1)=-3/2
y=-3/2x+b
find b
sub a point
(1,2)
2=-3/2(1)+b
2=-3/2+b
4/2=-3/2+b
7/2=b

y=-3/2x+7/2
3/2x+y=7/2
times 2 both sides
3x+2y=7 is standard form

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## CAN YOU PLEASE HELP ME WITH THIS PROBLEM? This is due by MIDNIGHT 2). Manufacturer A produces hammers that are normally distributed with a mean weight of 4.6 lb and a standard deviation of 0.8 lb. Manufacturer B produces hammers that are normally distributed with a mean weight of 6.3 lb and a standard deviation of 1.4 lb. (a) What percentage of Manufacturer A’s hammers will weigh less than 5 lb? (b) What percentage of Manufacturer B’s hammers will weigh less than 5 lb? (c) Which manufacturer is more likely to produce a hammer weighing exactly 5.2 lb? Explain.

Part (a)
Manufacturer A;
Mean, μ = 4.6
Std. deviation, σ = 0.8

For the random variable x = 5, the z-score is
z = (x – μ)/σ = (5 – 4.6)/0.8 = 0.5
From standard tables,
P(x<5) = P(z<0.5) = 0.69 = 69%

69% will weigh less than 5 lb.

Part b)
Manufacturer B
μ = 6.3
σ = 1.4

For x = 5, z = (5 – 6.3)/1.4 = -0.9286
From standard tables,
P(x<5) = P(z<-0.9286) = 0.1766 = 17.7%

About 18% will weigh less than 5 lb.

Part (c)
x = 5.2 lb

Manufacturer A:
z = (5.2-4.6)/0.8 =  0.75
From tables,
P(x=5.2) = 0.773 = 77.3%

Manufacturer B:
z = (5.2-6.3)/1.4 = -0.7857
P(x=5.2) = 0.216 = 21.6%

Manufacturer A is more likely to produce a 5.2 lb hammer because its probability is higher.

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## The body temperatures of adults are normally distributed with a mean of 98.6° f and a standard deviation of 0.60° f. if 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° f. 0.0228 0.8188 0.9360 0.9772

Here the population standard deviation is 0.60 degree F.  If a sample of 36 adults is randomly selected, that results in a sample standard deviation of 0.60 degree F divided by the square root of 36:  0.10 degree F.

The probability in question is the area under the standard normal probability distribution between 98.4 degree F and infinity, and intuitively you can detect that this will be more than 0.5 (corresponding to 50%).

Convert 98.4 degrees F to a z-score, using the sample standard deviation (0.10 degree F).  That z score is
98.4-98.6
z = ————–   =  -0.20/0.10 = -2
0.10

We need to determine the area under the standard normal curve to the right of z=-2.  Use a table of z-scores to do this, or use your calculator’s built-in probability functions.  My result is 98.21% (corresponding to an area of 0.9821).

With my calculator I can find this probability using the following command:

normalcdf(-2,100000,0.10).

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## When a standard pair of six sided dice are rolled, what is the probability of getting a 13?

When a standard pair of six sided dice are rolled, what is the probability of getting a 13?

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## A company is setting its direct materials and direct labor standards for its leading product. basic wages of the assembly line personnel are \$19 per hour. payroll taxes are approximately 23% of wages. how much is the direct labor cost standard per hour? (round your answer to the nearest cent.)

A company is setting its direct materials and direct labor standards for its leading product. basic wages of the assembly line personnel are \$19 per hour. payroll taxes are approximately 23% of wages. how much is the direct labor cost standard per hour? (round your answer to the nearest cent.)

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## Given that the random variable x is normally distributed with a mean of 80 and a standard deviation of 10, p(85 < x < 90) is

Given that the random variable x is normally distributed with a mean of 80 and a standard deviation of 10, p(85 < x < 90) is

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## A normal distribution of data has a mean of 15 and a standard deviation of 4. How many standard deviations from the mean is 25? 0.16 0.4 2.5 6.25

This is the concept of probability, to get the number of standard deviations that 25 is from the mean, we calculate the z-score given by:
Z=(X-mean)/s.d
where;
x=25
mean=15
s.d=4
hence;
z=(25-15)/4=2.5

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## Read the excerpt from “mother tongue.” lately, ive been giving more thought to the kind of english my mother speaks. like others, i have described it to people as broken or fractured english. but i wince when i say that. it has always bothered me that i can think of no other way to describe it other than broken, as if it were damaged and needed to be fixed, as if it lacked a certain wholeness and soundness. what best supports the inference that tan believes nonstandard english is no less valid than standard english? tan spends a lot of time thinking about her mothers fractured english. tan has trouble thinking of descriptive words when she is writing. tans american education makes it difficult for her to understand her mother. tan winces when she describes her mothers english as broken.

The following option best supports the inference that tan believes nonstandard English is no less valid than standard English :

Tan winces when she describes her mothers English as broken.

Tan winces because she considers her mother’s nonstandard English as her mother tongue. It is perfectly clear and natural to her. It is full of vivid descriptions, clear observations and imagery. Nonstandard English is close to her heart because it is the language that helped her to make sense of the world.

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## You are dealt one card from a standard 52-card deck. find the probability of being dealt an ace or a 9

(C)0.20

Step-by-step explanation:

We are given that we have to make five-letter codes that use the letters A, F, E, R, and M without repeating any letter.

Thus, total number of letters=5

Now, the probability that a randomly chosen code starts with A is given as:   Therefore, the the probability that a randomly chosen code starts with A is 0.20.

Hence, option C is correct.

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## Intel suggests it uses a standard discount rate across different lines of business. what would be a reasonable justification for intel to have this type of? policy

Initial investment on Jan 1, 2013 = (500 shares)*(\$24 per share) = \$12,000

Dividend collected at the end of 2013 = \$2.50*500 = \$1,250
Dividend collected at the end of 2014 = \$4*500 = \$2,000
Dividend collected at the end of 2015 = \$3*500 = \$1,500
Mony received from sellng the 500 shares at the end of 2015 = \$20*500 = \$10,000

Total returns at the end of 2015 = 1,250+2,000+1,500+10,000 = \$14,750
Net gains = 14750 – 12000 = \$2,750
Duration = 3 years
Realized total rate of return = 2750/12000 = 0.2292 = 22.9%