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Anastaziya [24]
3 years ago
10

Can someone help me with this geometry problem?

Mathematics
1 answer:
frozen [14]3 years ago
5 0
Answer: 35


Explanation:
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(x+x/2),(y+y/2) you have two x values and two y values to substitute into the equation
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The measure of each exterior angle of a regular octagon is the measure of each exterior angle of a regular hexagon
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I think its greater than

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3 years ago
How many people in a car? A study of rush-hour traffic in San Francisco counts the number of people in each car entering a freew
emmasim [6.3K]

Answer:

a) No. It is not normal.

b) The probability that 700 randomly selected cars at this freeway entrance will carry more than 1075 people is 0.104

Step-by-step explanation:

<u>(a) Could the exact distribution of the count be Normal?</u>

The exact distribution of the number of people in each car entering a freeway at a suburban interchange is not normal. Because the count is <em>discrete </em>and <em>can assume values bigger or equal to one</em>.

<u>(b) The probability that 700 randomly selected cars at this freeway entrance will carry more than 1075 people.</u>

The probability we seek is the cars carrying people with mean more than \frac{1075}{700}=1.5357

That is P(z>z*) where z* is the z-score of 1.5357.

z* can be calculated using the equation:

z*=\frac{X-M}{\frac{s}{\sqrt{N} } } where

  • X is the mean value wee seek for its z-score (1.5357)
  • M is the average count of people entering a freeway at a suburban interchange. (1.5)
  • s is the standard deviation of the count (0.75)
  • N is the sample size (700)

Thus z*=\frac{1.5357-1.5}{\frac{0.75}{\sqrt{700} } } ≈ 1.26

We have P(z>1.26)=1-P(z≤1.26)= 1-0.896 = 0.104

5 0
3 years ago
A recent survey found that 80.5% of households has Internet access and 82% of households had cable television. Also, it was repo
kirill115 [55]

Answer: 0.878

Step-by-step explanation:

Given : P( Internet access) = 80.5%= 0.805

P(cable television) = 82% = 0.82

P( Internet access and cable television) = 74.7% =0.747

Formula: P(A or B)= P(A ) + P(B) - P(A and B)

So, P( Internet access or cable television)  =  P( Internet access)  + P(cable television) -P( Internet access and cable television)

= 0.805+0.82-0.747

=  0.878

The probability that a randomly selected household in the survey had either Internet access or cable = 0.878

5 0
3 years ago
Solve. {2d+e=8d−e=4 Use the linear combination method.
OLga [1]
D=4 and e=0
hope it helps
4 0
4 years ago
Read 2 more answers
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