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saul85 [17]
3 years ago
5

he mean weight of a breed of yearling cattle is 11871187 pounds. Suppose that weights of all such animals can be described by th

e Normal model ​N(11871187​,7878​). ​a) How many standard deviations from the mean would a steer weighing 10001000 pounds​ be? ​b) Which would be more​ unusual, a steer weighing 10001000 ​pounds, or one weighing 12501250 ​pounds? ​a) A steer weighing 10001000 pounds is nothing standard deviations below the mean.
Engineering
1 answer:
Tamiku [17]3 years ago
8 0

Question

The mean weight of a breed of yearling cattle is 1187 pounds. Suppose that weights of all such animals can be described by the Normal model ​N(1187,78). ​

a) How many standard deviations from the mean would a steer weighing 1000 pounds​ be?

b) Which would be more​ unusual, a steer weighing 1000 ​pounds, or one weighing 1250 ​pounds? ​

Answer:

a. z = -2.40

A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.

b. z = 0.81

1000 is more unusual because its contained on the extreme end from the mean

Explanation:

a.

Let weight (in pounds) of the cattle be denoted by letter x:

z = (x - u)/ σ

Where u = mean and σ = standard deviation

u = 1187

σ = 78

x = 1000

Use z score formula to standardize the value of x:

z = (1000 - 1187)/78

z = -187/78

z = -2.397436

z = -2.40 ------_ Approximated

A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.

b.

x= 1250

z= (1250 - 1187)/78

z = 63/78

z = 0.807692

z = 0.81 --------- Approximated

1000 is more unusual because its contained on the extreme end from the mean

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Answer: ALL CAREFULLY ANSWERED CORRECTLY.

Explanation:

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7 0
3 years ago
A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2
IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length l for the deformation, we use the following relation;

l = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

l = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

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5 0
3 years ago
Draw the ipo chart for a program that reads a number from the user and display the square of that number ???Anyone please
kompoz [17]

Answer:

See attachment for chart

Explanation:

The IPO chart implements he following algorithm

The expressions in bracket are typical examples

<u>Input</u>

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<u>Processing</u>

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Nina [5.8K]

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6 0
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dsp73

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<h3>What are thematic groups?</h3>

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The control of dynamic system in order to produce the desirable outcome has been the role of the control system.

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Learn more about thematic group, here:

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