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damaskus [11]
1 year ago
8

Please help!!! i need this done by tonight!! 50 points!!

Mathematics
1 answer:
Vinil7 [7]1 year ago
5 0

The equation of best fit is y = (-22/5)x + 10.

What is an equation of a line?

The equation of a line is given by:

y = mx + c where m is the slope of the line and c is the y-intercept.

Example:

The slope of the line y = 2x + 3 is 2.

The slope of a line that passes through (1, 2) and (2, 3) is 1.

We have,

The following coordinates are given:

Pick two coordinates.

(0, 10) and (25, -100)

The equation of best fit.

y = mx + c

Now,

m = (-100 - 10) / (25 - 0)

m = -110 / 25

m = -22/5

Now,

(0, 10) = (x, y)

10 = (-22/5) x 0 + c

c = 10

Now,

y = mx + c

y = (-22/5)x + 10

Thus,

Using the coordinates the equation of best fit is y = (-22/5)x + 10.

Learn more about equation of a line here:

brainly.com/question/23087740

#SPJ1

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Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

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

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

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

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

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

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

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

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
3 years ago
Chris and his brother leave the same point and run in opposite directions on a straight road. After running for 3 Hours they are
ikadub [295]
<span>Chris' average speed is 3.5mph: Lets take Chris' brother speed = B then Chris speed = (B+ 1) mph-------given in the question and: The average speed for covering 18 miles= chris' average speed + his brother's average speed = (B+ 1) + B since Speed = Distance covered / time taken then average speed for 18 miles can also be found by: 18 miles/3 hours = 6 miles per hour then we have; (B+ 1) + B = 6mph (B +1) = (6-B) B+B = 6-1=5 2B=5 B=5/2=2.5 mph (this is Chris' brother average speed) Hence Chris' average speed is = B+1= 2.5+1= 3.5mph</span>
8 0
3 years ago
Randy and Carla like to walk the path
Maksim231197 [3]
<h3>Answer:  18 miles</h3>

Reason:

The path is 2 miles and Randy walked it 13 times. He walked a total of 2*13 = 26 miles.

Carla walked the path 22 times, so she walked a total of 2*22 = 44 miles.

The difference of those results is 44-26 = 18 miles.

If you wanted to do the full calculation in one line, then it would be

2*22 - 2*13 = 18

or you could say

2*(22-13) = 18

8 0
2 years ago
Help!<br> Change 2/5<br> to a percent.
nalin [4]

Answer:

40%

Step-by-step explanation:

6 0
3 years ago
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Twenty people are going by van to a movie. Each van seats 8 people. How many vans are needed to take everyone?
serg [7]
Yes,you are correct. 
The first van carries 8 people.
The second van carries 8 people. 
The 3rd van carries the remaining 4 people. 
8+8+4=20
7 0
4 years ago
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