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kirill [66]
3 years ago
6

H(8) = 40,000 • 1.05^8

Mathematics
1 answer:
katovenus [111]3 years ago
3 0

Answer:

height =8 = 40,000*1.0

decide to do it doesn't matter to me I am not earlier then I can also need to get income tax rate in the us

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Find the equation of the line that is parallel to the given line and passes through (9,1) y=2/3x -2
Nadya [2.5K]
When the lines are parallel, the slopes of these lines are equal.
For a  new line,
y = (2/3)x + b.

To find b, we need substitute x and y with values of the point (9,1).
x=9
y=1

1 = (2/3)*9 + b
1 = 6 + b
b=-5

The equation of the line is y = (2/3)x - 5.



8 0
2 years ago
Clarence works at least 5 hours but not more than 7 hours. he earns $11.60 per hour. the function f(t)=11.6t represents the amou
Pavel [41]
Function: f(t) = 11.6 t

Restrictions: 5 ≤ t ≤ 7

Domain: possible values of t: all real values between 5 and 7 inclusive

Range: possible values of f(t): from 11.6 * 5 to 11.6 *7 = from 58 to 81.2, inclusive.

Answer:

the practical domain is all real numbers from 5 to 7, inclusive. and the practical range is all real numbers from 58 to 81.2

4 0
3 years ago
Read 2 more answers
Please help me with this problem. Do not send me a link to some random website. I do not trust the link given that requires me t
krok68 [10]

Answer: \left[\begin{array}{ccc}|x|0|1|2|3|4|5|6|\\|P(x)|0|4|8|12|16|20|24|\end{array}\right]

b. P(x)= 4x

c. The graph would go in a straight line touching each point given.

Step-by-step explanation: To find the perimeter of a square you can use the equation 4(x). If you multiply each x value by 4 you'll get the P(x) values.

3 0
2 years ago
The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar
Margarita [4]

Answer:

  1. 0.26
  2. 0.91
  3. 1.43

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given data

mean = 1.9 hours

standard deviation = 0.3 hours

solution

we get here first  random movie between 1.8 and 2.0 hours

so here

P(1.8 < z < 2 )

z = (1.8 - 1.9) ÷ 0.3

z = -0.33

and

z = (2.0 - 1.9) ÷ 0.3

z = 0.33

z = 0.6293

so

P(-0.333 < z < 0.333 )

=  0.26

so random movie is between 1.8 and 2.0 hours long is 0.26

and

A movie is longer than 2.3 hours.

P(x > 2.3)

P( \frac{x-\mu }{\sigma}  > \frac{2.3-\mu }{\sigma} )

P (z  > \frac{2.3-1.9 }{0.3}  )

P (z  > 1.333  )

= 0.091

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and

length of movie that is shorter than 94% of the movies is

P(x > a ) = 0.94

P(x <  a ) = 0.06

so

P( \frac{x-\mu }{\sigma } <  \frac{a-\mu }{\sigma } )

\frac{a-1.9 }{0.3 } = -1.55

a = 1.43

so length of the movie that is shorter than 94% of the movies about 1.4 hours.

3 0
3 years ago
I need help with this . I don’t get how
Svetradugi [14.3K]
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For example 15x^2 and 10x^2 are like terms. Circle the sign in front of the term to perform the correct operation.







5 0
3 years ago
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