All eight companies in the aerospace industry were surveyed as to their return on investment last year. The results are in percent) 10.6, 12.6, 14.8, 18.2, 12.0, 14.8, 12.2, and 15.6. a. Calculate the range. (Round your answer to 1 decimal place.) Range %
b. Calculate the arithmetic mean. (Round your answer decimal places.)
c. Calculate the variance. (Round your answer to 2 decimal places.)

Answers

Answer 1
Answer:

Answer:

a.) range- 6.2

b.) arithmetic mean- 13.85

c.) varience- 0.69


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you order two tacos and a salad. The salad cost $2.50. You leave a three-dollar tip. Theres is no sales tax. How much does one taco cost? Show your work.

Answers

Answer:

$5.50

Step-by-step explanation:

2.50+3.00=5.50

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Find the area of the right triangle.

Answers

Answer:

area is 28 units²

Step-by-step explanation:

find the area of the rectangle with sides 7 and 8 and divide it in half

A=7*8/2= 56/2=28

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with μ = 44 and σ = 5.0. (a) What is the probability that yield strength is at most 40? Greater than 62? (Round your answers to four decimal places.) at most 40 greater than 62 (b) What yield strength value separates the strongest 75% from the others? (Round your answer to three decimal places.)

Answers

Answer:

Step-by-step explanation:

Using normal distribution,

z = (x - μ)/σ

μ= mean = 44 and

σ = standard deviation= 5.0

a) The probability that yield strength is at most 40=

P( x lesser than or equal to 40)

z = (40-44)/5= -0.8

Looking at the normal distribution table,

P( x lesser than or equal to 40) =0.2119

b) P(x greater than 62) = 1 - P(x lesser than or equal to 62)

z = (62-44)/5= 3.6

Looking at the normal distribution table,

P(x greater than 62) = 1 -0.99984

= 0.00016

c)P( 42 lesser than or equal to x lesser than or equal to 62)

= P(x lesser than or equal to 62) - P( x lesser than or equal to 40)

= 0.99987-0.2119= 0.78797

d) What yield strength value separates the strongest 75% from the others.

To get x for strongest 75, we get the z value corresponding to 0.75 from the table

z = 0.675= (x-44)/5

x = 3.375+44 = 47.375

The rest is 25% = 0.25

we get the z value corresponding to 0.25 from the table)

z = -0.67 = (x-44)/5

-3.35= x -44

x = -44+3.35= 40.65

yield strength value that separates the strongest 75% from the others

=47.375-40.65= 6.725

Final answer:

The probability that the yield strength is at most 40 is approximately 0.2119 and the probability that it is greater than 62 is approximately 0.0001. The yield strength value that separates the strongest 75% from the others is approximately 40.628 ksi.

Explanation:

This question is about calculating probabilities and percentiles using the properties of the normal distribution. The yield strength for the A36 grade steel is normally distributed with a mean (μ) of 44 and a standard deviation (σ) of 5.0.

(a) To find the probability that the yield strength is at most 40, we will need to calculate the Z-score value for the yield strength of 40. The Z-score can be calculated using the following formula: Z = (X - μ) / σ , where X is the observed value, μ is the mean, and σ is the standard deviation. For X = 40, μ = 44, and σ = 5.0, the Z-score is -0.8. Looking up the Z-score in the standard normal distribution table, the probability that the yield strength is at most 40 is approximately 0.2119. Using a similar process, we find that the probability that the yield strength is greater than 62 is less than 0.0001, very close to zero.

(b) To determine the yield strength value that separates the strongest 75% from the others, we find the Z-score that corresponds to a cumulative probability of 0.25 in the standard normal distribution table (because the strongest 75% corresponds to the weakest 25%). That Z-score is approximately -0.6745. Using the formula Z = (X - μ) / σ to solve for X gives us X = σZ + μ  = 5.0 * -0.6745 + 44 = 40.6275, which rounded to three decimal places is 40.628.

Learn more about Normal Distribution here:

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Compute 17÷2 enter your answer using remainder notation​

Answers

Final answer:

To compute 17 ÷ 2 using remainder notation, divide 17 by 2 and find the quotient and remainder. The quotient is 8 and the remainder is 1.

Explanation:

To compute 17 ÷ 2 using remainder notation, you divide 17 by 2 and find the quotient and remainder. In this case, the quotient is the whole number part of the division and the remainder is the leftover part. When you divide 17 by 2, the quotient is 8 and the remainder is 1. Therefore, the answer is 8 with a remainder of 1.

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Answer:8.5

Step-by-step explanation:

A line passes through A(3,7) and B(-4,9). Find the value of a if C(a, 1) is on the line.

Answers

Answer:  a=24

Step-by-step explanation:

Lets find the line's  formula (equation of the line).

As known the general formula of any straight line (linear function) is

y=kx+b

Lets find the coefficient k= (Yb-Ya)/(Xb-Xa)=(9-7)/(-4-3)=-2/7

(Xb;Yb)- are the coordinates of point B

(Xa;Ya) are the coordinates of point A

Now lets find the coefficient b.  For this purpose we gonna use the coordinates of any point A or B.

We will use A

7=-2/7*3+b

7=-6/7+b

b=7 6/7

So the line' s  equation is y= -2/7*x+7 6/7

Now we gonna find the value of a usingcoordinates of point C.

Yc=1,  Xc=a

1=-2/7*a+7 6/7

2/7*a= 7 6/7-1

2/7*a=6 6/7

(2/7)*a=48/7

a=48/7: (2/7)

a=24

Answer:

a=24

Step-by-step explanation:

The use of social networks has grown dramatically all over the world. In a recent sample of 24 American social network users and each was asked for the amount of time spent (in hours) social networking each day. The mean time spent was 3.19 hours with a standard deviation of 0.2903 hours. Find a 99% confidence interval for the true mean amount of time Americans spend social networking each day

Answers

Answer:

The 99% confidence interval for the true mean amount of time Americans spend social networking each day is (3.02 hours, 3.36 hours).

Step-by-step explanation:

The (1 - α)% confidence interval for population mean when the population standard deviation is not known is:

CI=\bar x\pm t_(\alpha/2, (n-1))* (s)/(√(n))

The information provided is:

n=24\n\bar x=3.19\ \text{hours}\ns=0.2903\ \text{hours}

Confidence level = 99%.

Compute the critical value of t for 99% confidence interval and (n - 1) degrees of freedom as follows:

t_(\alpha/2, (n-1))=t_(0.01/2, (24-1))=t_(0.005, 23)=2.807

*Use a t-table.

Compute the 99% confidence interval for the true mean amount of time Americans spend social networking each day as follows:

CI=\bar x\pm t_(\alpha/2, (n-1))* (s)/(√(n))

     =3.19\pm 2.807* (0.2903)/(√(24))\n\n=3.19\pm 0.1663\n\n=(3.0237, 3.3563)\n\n\approx (3.02, 3.36)

Thus, the 99% confidence interval for the true mean amount of time Americans spend social networking each day is (3.02 hours, 3.36 hours).