Answer:
55
Step-by-step explanation:
The equation of the line is y = -11x + 232
<h3>How to determine the equation?</h3>
The given parameters are:
Slope (m)= -11
Point (x1, y1) = (31, -109)
The linear equation is then calculated as:
y = m(x - x1) + y1
This gives
y = -11(x - 31) - 109
Evaluate the product
y = -11x + 341 - 109
Evaluate the like terms
y = -11x + 232
Hence, the equation of the line is y = -11x + 232
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Answer:
D
Step-by-step explanation:
100+25 * 12 ; 100+300 = 400 ; difference = 400-320=80 ; percent increase since it is 1 year = 80/320 =1/4=0.25 = 25/100=25%
Answer:
95% Confidence interval = (23.4,26.2)
Step-by-step explanation:
In this problem we have to develop a 95% CI for the mean.
The sample size is n=49, the mean of the sample is M=24.8 and the standard deviation of the population is σ=5.
We know that for a 95% CI, the z-value is 1.96.
The CI is
Answer: If we define 2:00pm as our 0 in time; then:
at t= 0. the velocity is 30 mi/h.
then at t = 10m (or 1/6 hours) the velocity is 50mi/h
Then, if we think in the "mean acceleration" as the slope between the two velocities, we can find the slope as:
a= (y2 - y1)/(x2 - x1) = (50 mi/h - 30 mi/h)/(1/6h - 0h) = 20*6mi/(h*h) = 120mi/
Now, this is the slope of the mean acceleration between t= 0h and t = 1/6h, then we can use the mean value theorem; who says that if F is a differentiable function on the interval (a,b), then exist at least one point c between a and b where F'(c) = (F(b) - F(a))/(b - a)
So if v is differentiable, then there is a time T between 0h and 1/6h where v(T) = 120mi/