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Gre4nikov [31]
3 years ago
6

The actual length of a soccer field is 300 feet. If the width of the field is 180 feet, then what should the width of the scale

drawing be in inches?
Mathematics
1 answer:
topjm [15]3 years ago
6 0

Answer:

2160 inches

Step-by-step explanation:

The actual length of a soccer field is 300 feet. If the width of the field is 180 feet, then what should the width of the scale drawing be in inches?

1 foot = 12 inches

Width of the field = 180 feet

Hence:

1 foot = 12 inches

180 feet = x

x = 180 × 12 inches

x = 2160 inches

Width of the scale drawing in inches = 2160 inches

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Domain and Range for the function f(x)=5IXI is
shutvik [7]

Answer:

The domain of the function f(x) is:

\mathrm{Domain\:of\:}\:5\left|x\right|\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

The range of the function f(x) is:

\mathrm{Range\:of\:}5\left|x\right|:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

Step-by-step explanation:

Given the function

f\left(x\right)=5\left|x\right|

Determining the domain:

We know that the domain of the function is the set of input or arguments for which the function is real and defined.  

In other words,  

  • Domain refers to all the possible sets of input values on the x-axis.

It is clear that the function has undefined points nor domain constraints.

Thus, the domain of the function f(x) is:

\mathrm{Domain\:of\:}\:5\left|x\right|\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

Determining the range:

We also know that range is the set of values of the dependent variable for which a function is defined.  

In other words,  

  • Range refers to all the possible sets of output values on the y-axis.

We know that the range of an Absolute function is of the form

c|ax+b|+k\:\mathrm{is}\:\:f\left(x\right)\ge \:k

k=0

so

Thus, the range of the function f(x) is:

\mathrm{Range\:of\:}5\left|x\right|:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

7 0
3 years ago
The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation 5 pounds. If 50,000 part
nalin [4]

Answer:

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

7933 parts

b) How many would have a tensile strength in excess of 48 pounds?

2739.95 parts

Step-by-step explanation:

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

z = (x-μ)/σ

x = 35 μ = 40 , σ = 5

z = 35 - 40/5

= -5/-5

= -1

Determining the Probability value from Z-Table:

P(x<35) = 0.15866

Converting to percentage = 15.866%

We are asked how many will fail to meet this specification

We have 50,000 parts

Hence,

15.866% of 50,000 parts will fail to meet the specification

= 15.866% of 50,000

= 7933 parts

Therefore, 7933 parts will fail to meet the specifications.

b) How many would have a tensile strength in excess of 48 pounds?

z = (x-μ)/σ

x = 48 μ = 40 , σ = 5

z = 48 - 40/5

z = 8/5

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P(x>48) = 1 - P(x<48)

1 - 0.9452

= 0.054799

Converting to percentage

= 5.4799%

Therefore, 5.4799% will have an excess of (or will be greater than) 48 pounds

We are asked, how many would have a tensile strength in excess of 48 pounds?

This would be 5.4799% of 50,000 parts

= 5.4799% × 50,000

= 2739.95

Therefore, 2739.95 parts will have a tensile strength excess of 48 pounds

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