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nikklg [1K]
3 years ago
7

What are 3 x-intercepts of f(x)=x^2-8x+7

Mathematics
1 answer:
Nadya [2.5K]3 years ago
7 0

Answer:

you use to computer:

x1=7

x2=1

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What is the value of each expression in simplest form? 9/1 divided by 3/5
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9/1 is simply 9
9 divided by 3/5 = 15
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4 years ago
Solve for x: 21x - 3| + 1 = 7
Andreas93 [3]

Answer:

x=6

Step-by-step explanation:

distribute 2 to x and 3..you get 2x -6 +1. add 6 and subtract one to cancel it out and bring it to the other side. 2x=12...simplify x=6

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Just number 7, please help quickly I don’t know what I’m doing (if you give an explanation I’ll mark as brainliest)
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Answer: b

Step-by-step explanation:

x=12

12+5=17

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10ml would be the answer to this question
7 0
3 years ago
The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. However,
horsena [70]

Answer:

The mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

Step-by-step explanation:

We are given that records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.

On a typical​ Saturday, the​ oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Assuming that data follows normal distribution.

Let \bar X = <u><em>sample mean time</em></u>

The z score probability distribution for sample mean is given by;

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

where, \mu = population mean time = 16.2 minutes

            \sigma = standard deviation = 3.4 minutes

            n = sample size = 35 oil changes

<u>Now, the mean oil-change time that would there be a 10% chance of being at or below is given by;</u>

<u></u>

          P(X \leq x) = 0.10          {where x is required mean oil-change time}

          P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

          P(Z \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

Now, in the z table the critical value of X which represents the below 10% of the probability area is given as -1.282, that means;

                  \frac{x-16.2}{\frac{3.4}{\sqrt{35} } }  =  -1.282

               x - 16.2  =  -1.282 \times {\frac{3.4}{\sqrt{35} } }

                         x  =  16.2 - 0.74 = <u>15.46 minutes</u>

Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

8 0
3 years ago
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