Answer:
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
Step-by-step explanation:
We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.
We have to calculate a 95% confidence interval for the proportion.
The sample proportion is p=0.26.
The standard error of the proportion is:
The critical z-value for a 95% confidence interval is z=1.96.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:
The 95% confidence interval for the population proportion is (0.192, 0.328).
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
8
46
is equivalent to 4
23
because 4 x 2 = 8 and 23 x 2 = 46
12
69
is equivalent to 4
23
because 4 x 3 = 12 and 23 x 3 = 69
16
92
is equivalent to 4
23
because 4 x 4 = 16 and 23 x 4 = 92
It will take her 8 weeks to have enough money for a bike. I made the equation 125+15x=245 and multiplied 15 by 8 to get exactly 245
Answer:
a. 54.05 Mpbs.
b. 2.745... standard deviations.
c. The z-score is 2.745....
d. The carrier's highest data speed is significantly high.
Step-by-step explanation:
a. The difference between the highest measured data speed and the mean is 72.6 - 18.55 = 54.05 Mbps.
b. The amount of standard deviations of 54.05 Mbps is equal to this value divided by the standard deviations, so we yield 
standard deviations.
c. The z-score is equal to the difference between the mean and a data point in standard deviations, so the z-score is 2.745....
d. 2.745... is not between -2 and 2, so the carrier's highest data speed is not insignificant - so it's significantly high.
120 - 3/4y=60
lowest common multiple=4y
480y-3=240y
480y-240y=3
240y=3
y=3/240
y=1/80
Answer: y=1/80
