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3 years ago

Give an example of a function in everyday life and represent it

1 answer:

5 0

Wake up getting dressed (I’m a cylindrical metal fusion tech Going to my line of work requires work as fire resistance clothing and quality work
Boots , kiss my wife over and over and over and tell her I love her and check on the kids before I leave then pull My 7’12’sshift . Drive home ,walk in the door And kiss my beautiful wife who is worked hard all day slaving herself around the house sit down eat my supper that she brings me play with the kids take a shower go to bed and repeat step one

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PLZ HELP IN A RUSH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Bumek [7]

Answer:

Her new balance = - 92.34

Step-by-step explanation:

First find out what the check did.

She had $63.44 in her account. Her check caused her to be in the whole

63.44 - 186.56 = -123.12

Now she deposited 1/4 of that amount.

- 123.12 + (1/4)(123.12) = - 123.12 + 30.78

Her new balance = - 92.34

4 0

3 years ago

Alex73 [517]

Answer:

2x + 7= 4x + 17

7=2x+17

-10=2x

-10/2=2x/2

x=−5

5 0

2 years ago

Graph the inequality and check the solution.

denis-greek [22]

Answer:

B,C,E

Step-by-step explanation:

did the assignment on Edge 2020

B. draw an open circle at 100

C. shade all numbers to the right of 100

E. 150 is shaded, so substitute 150 in for the variable to check the graph of the solution

5 0

2 years ago

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l

tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

Step(ii):-

The Empirical rule

$Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }$

Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = P(Z≥2.383)

= 1- P( Z<2.383)

= 1-( 0.5 -+A(2.38))

= 0.5 - A(2.38)

= 0.5 -0.4913

= 0.0087

Final answer:-

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

3 0

2 years ago

Test 69,000 on hearing to for divisibility by 2 3 5 9 or 10

svetoff [14.1K]

Yes they are correct

7 0

3 years ago