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SOVA2 [1]
1 year ago
9

The length of a school bus is 12.6meters.if 9 school buses park end to end with 2meters between each one, what’s the total lengt

h from the front of the first bus to the end of the last bus?
Mathematics
2 answers:
marta [7]1 year ago
7 0

First, 12.6 times 9 is 113.4. So withouth the space, that is how long the busses are. Then you do 2meters times 8 is 16 meters between the busses. 113.4+16 equals 129.4, the final answer.

Masja [62]1 year ago
7 0

Answer:  The total length from the front of the first bus to the end of the last bus is 129.4 meters.

Step-by-step explanation:

Hi, to answer this question we have to multiply:

The number of buses by the length of each one

  • 12.6 x 9 = 113.4 meters

The number of spaces between buses by the length of each space.

Since there is no space in front of the first bus and behind the last bus, there are 8 spaces in between.

  • 8 x 2 = 16
  • Adding both results: 16 + 113.4 =129.4 meters front to back

Feel free to ask for more if needed or if you did not understand something.  

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The weekly salary paid to employees of a small company that supplies​ part-time laborers averages ​$750 with a standard deviatio
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Answer:

(a) The fraction of employees is 0.84.

(b)

\mu=850\\\\\sigma=450

(c)

\mu=787.5\\\\\sigma=472.5

(d) No. The left part of the distribution would be truncated too much.

Step-by-step explanation:

(a) If the weekly salaries are normally​ distributed, estimate the fraction of employees that make more than ​$300 per week.

We have to calculate the z-value and compute the probability

z=\frac{X-\mu}{\sigma}= \frac{300-750}{450}=\frac{-450}{450}=-1\\\\P(X>300)=P(z>-1)=0.84

(b) If every employee receives a​ year-end bonus that adds ​$100 to the paycheck in the final​ week, how does this change the normal model for that​ week?

The mean of the salaries grows $100.

\mu_{new}=E(x+C)=E(x)+E(C)=\mu+C=750+100=850

The standard deviation stays the same ($450)

\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(x+C)-(\mu+C)]^2}  } =\sqrt{\frac{1}{N} \sum{(x+C-\mu-C)^2}  }\\\\ \sigma_{new}=\sqrt{\frac{1}{N} \sum{(x-\mu)^2}  } =\sigma

(c) If every employee receives a 5​% salary increase for the next​ year, how does the normal model​ change?

The increases means a salary X is multiplied by 1.05 (1.05X)

The mean of the salaries grows 5%, to $787.5.

\mu_{new}=E(ax)=a*E(x)=a*\mu=1.05*750=487.5

The standard deviation increases by a 5% ($472.5)

\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(ax)-(a\mu)]^2}  } =\sqrt{\frac{1}{N} \sum{a^2(x-\mu)^2}  }\\\\ \sigma_{new}=\sqrt{a^2}\sqrt{\frac{1}{N} \sum{(x-\mu)^2}}=a*\sigma=1.05*450=472.5

(d) If the lowest salary is ​$300 and the median salary is ​$525​, does a normal model appear​ appropriate?

No. The left part of the distribution would be truncated too much.

7 0
1 year ago
is x + 10 a factor of the function f(x) = x3 − 75x + 250? Explain. (2 points) Yes. When the function f(x) = x3 − 75x + 250 is di
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Asked and answered elsewhere.
brainly.com/question/10474342

Your appropriate choice is ...
  Yes. When the function f(x) = x3 − 75x + 250 is divided by x + 10, the remainder is zero. Therefore, x + 10 is a factor of f(x) = x3 − 75x + 250.
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Answer: a) Length of wire to replace old one=9ft

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Step-by-step explanation: from the right angle triangle, Tan&=20/15

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Length of wire,x=cos 31.13=x/15

15cos31.13

X=9

b) Tan&=10/6

Tan^-110/6

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8 0
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