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notka56 [123]
3 years ago
12

Need answer as fast as possible

Mathematics
1 answer:
jekas [21]3 years ago
3 0
Answer:
No solution

Explanation:
Because the lines are parallel which men’s story not touching at all no contact at all. Therefore if they were crossing they would have ONE solution. And so if u see just one line it’s not one line it’s two but u can’t see it because they are lined up together therefor Ethan would be infinite solutions
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i tihnk it is = − 4 im sorry if it is wrong

Step-by-step explanation:

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400,000 + 40,000????
Ilia_Sergeevich [38]

Answer:

440,000!!! LOL

Step-by-step explanation:

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Lorico [155]
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4 years ago
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A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that the mean
Misha Larkins [42]

Answer: 0.9608

Step-by-step explanation:

Given : A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that

Mean :\mu=8.4\text{ hours}

Standard deviation : \sigma=1.8\text{ hours}

Sample size : n=40

Let \overline{x} be the sample mean.

The formula for z-score in a normal distribution :

z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For  \overline{x}=8.9

z=\dfrac{8.9-8.4}{\dfrac{1.8}{\sqrt{40}}}\approx1.76

The P-value = P(\overline{x}

Hence, the probability that their mean rebuild time is less than 8.9 hours is 0.9608 .

7 0
3 years ago
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The number of chips in a bag is normally distributed with a mean of 74 and a standard deviation of 4. Approximately what percent
pav-90 [236]

Answer:

99.73% of bags contain between 62 and 86 chips .

Step-by-step explanation:

We are given that the number of chips in a bag is normally distributed with a mean of 74 and a standard deviation of 4.

Let X = percent of bags containing chips

So, X ~ N(\mu = 74 , \sigma^{2}=4^{2})

The standard normal z score distribution is given by;

         Z = \frac{X-\mu}{\sigma} ~ N(0,1)

So, percent of bags contain between 62 and 86 chips is given by;  

P(62 < X < 86) = P(X < 86) - P(X <= 62)                              

P(X < 86) = P( \frac{X-\mu}{\sigma} < \frac{86-74}{4} ) = P(Z < 3) = 0.99865 {using z table}

P(X <= 62) = P( \frac{X-\mu}{\sigma} <= \frac{62-74}{4} ) = P(Z <= -3) = 1 - P(Z < 3)= 1 - 0.99865 = 0.00135

So, P(62 < X < 86) = 0.99865 - 0.00135 = 0.9973 or 99.73%

Therefore, 99.73% of bags contain between 62 and 86 chips .

                                                                               

8 0
4 years ago
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