Answer:
x=4 x = 3/2
Step-by-step explanation:
3(x – 4)(2x - 3) = 0
We can use the zero product property
x-4 =0 2x-3=0
Solve each equation
x-4 =0 2x-3=0
x=4 2x=3
x =3/2
24.8 < μ < 27.0
24.2 < μ < 27.6
24.4 < μ < 27.4
The 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service is;
Option A; 24.6 < μ < 27.2
We are given;
Sample mean; x' = 25.9 pounds
Standard deviation; σ = 3.8 pounds
Confidence level; CL = 95%
Sample size; n = 34
Now formula for confidence interval is;
CI = x' ± z( σ/√n)
Where z is critical value at given confidence level.
z at CL of 95% is; z = 1.96
Thus;
CI = 25.9 ± 1.96(3.8/√34)
CI = 25.9 ± 1.3
CI = (25.9 - 1.3), (25.9 + 1.3)
CI = (24.6, 27.2)
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Answer:
24.6 < μ < 27.2
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 25.9 - 1.3 = 24.6 pounds.
The upper end of the interval is the sample mean added to M. So it is 25.9 + 1.3 = 27.2 pounds.
So the correct answer is:
24.6 < μ < 27.2
Answer:
r3523w5r23w5
Step-by-step explanation:
a.S= 230T– 110; 1500 lbc.T= 230S– 110; 1500 lbb.S= 230T+ 110; 1720 lbd.S= 199T– 110; 1280 lba.S= 230T– 110; 1500 lbc.T= 230S– 110; 1500 lbb.S= 230T+ 110; 1720 lbd.S= 199T– 110; 1280 lba.S= 230T– 110; 1500 lbc.T= 230S– 110; 1500 lbb.S= 230T+ 110; 1720 lbd.S= 199T– 110; 1280 lba.S= 230T– 110; 1500 lbc.T= 230S– 110; 1500 lbb.S= 230T+ 110; 1720 lbd.S= 199T– 110; 1280 lba.S= 230T– 110; 1500 lbc.T= 230S– 110; 1500 lbb.S= 230T+ 110; 1720 lbd.S= 199T– 110; 1280 lb