Given that,
Sample size= 83
Mean number= 39.04
Standard deviation= 11.51
We know the critical t-value for 95% confidence interval which is equal to 1.989.
We also know the formula for confidence interval,
CI=( mean number - critical t-value*standard deviation/(sample size)^(1/2), mean number + critical t-value*standard deviation/(sample size)^(1/2))
So, we have
CI= (39.04 - 1.989*11.51/83^(1/2), 39.04 + 1.989*11.51/83^(1/2)
CI= (39.04 - 2.513,39.04 + 2.513)
CI= (36.527,41.553)
Therefore, 95% confidence interval for these data is (36.527,41.553), and this result interpret that the true value for this survey sample lie in the interval (36.527,41.553).
Answer:
slope = -1/3, y-intercept (0, -3), x-intercept (-9 , 0)
Step-by-step explanation:
3 - x = 3(y + 4)
3 - x = 3y + 12
3y = 3 - x - 12
3y = -x -9
y = -1/3 x - 3
slope = -1/3
when x = 0 (y-intercept)
y = -1/3 × 0 - 3
y = -3
y-intercept (0, -3)
when y = 0 (x-intercept)
0 = -1/3 x - 3
1/3 x = -3
x = -9
x-intercept (-9 , 0)
50,000 + 100x = 75,000
He sold 250 Computers last year
