FREE BRAINLIEST!! ill also answer questions that you have posted if you answer correctly!!!! (61pts)

Mathematics

1 answer:

Evgesh-ka [11]2 years ago

8 0

Answer:

Can I get the pdf?

Step-by-step explanation:

You might be interested in

A line passes through $A(1,1)$ and $B(100,1000)$. How many other points with integer coordinates are on the line and strictly be

jonny [76]

Answer:

Step-by-step explanation:

Integers are whole numbers or opposite of whole numbers.

The slope is 1.

How?

Change of y is 100-1=99.

Change of x is 100-1=99.

The slope is 99/99=1 or 1/1.

So if we start at (1,1) and we rise 1 and run 1 right, we get (2,2).

If we do that again from (2,2) we get (3,3).

Following the pattern of going up 1 and right from each new location discovered we get all the of these points:

Start (1,1)

(2,2)

(3,3)

(4,4)

(5,5)

(6,6)

....

(96,96)

(97,97)

(98,98)

(99,99)

end (100,100)

So we just need to count all the numbers from 2 to 99.

99-2+1

97+1

7 0

2 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is: