Please check answer!

15 There are 30 students in a class. 1 5 like football. How many students do not like football?

almond37 [142]

15

Explanation:
30-15= 15

5 0

2 years ago

I really don't get these answers but if you could help me that'll mean a lot thank you for answering my question if you did

mixer [17]

Use KMF (Keep multiply flip)
keep the first fraction as it is
change sign to multiplication
flip numerator and denominator of last fraction
then finish

6 0

3 years ago

Using the given zero, find one other zero of f(x). i is a zero of f(x).= x4 - 2x3 + 38x2 - 2x + 37.

statuscvo [17]

If i is a zero, -i is also a zero.
The way you get a complex zero is by taking the square root of a negative number. Taking the sqrt (-1) = + and - i

4 0

3 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: