Slope of first graph (-3+8)/(-5+10) = 5/5 = 1
Slope of second graph = 4
Slope of Function A is less than that of Function B
F = M/t, where M = mv so...
F = (42*0.65)/(0.018) = 1516.7 N
To the correct number of sig figs it is 1500 N
The answer is x=4/3 and x=11/4
Answer:
Her first coupon ti be used is $50 off a purchase above $300
Final purchase price= $340
Step-by-step explanation:
Marie has two coupons
one for a 15% discount and one for $50 off any purchase above $300.
The stores allow the two coupons to be combined and she spends a total of $450.
Her first coupon ti be used is $50 off a purchase above $300.
So Marie have $400 now
15% off $400= 400-(0.15*400)
15% off $400 = 400-60
15% off $400 =$ 340
Answer:
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.
This means that ![n = 1067, \pi = \frac{74}{1067} = 0.069](https://tex.z-dn.net/?f=n%20%3D%201067%2C%20%5Cpi%20%3D%20%5Cfrac%7B74%7D%7B1067%7D%20%3D%200.069)
80% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 - 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.059](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.069%20-%201.28%5Csqrt%7B%5Cfrac%7B0.069%2A0.931%7D%7B1067%7D%7D%20%3D%200.059)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 + 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.079](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.069%20%2B%201.28%5Csqrt%7B%5Cfrac%7B0.069%2A0.931%7D%7B1067%7D%7D%20%3D%200.079)
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).