Answer: 90 degrees
Step-by-step explanation:
Given:
Area of sample board: 1/3 square ft.
Weight of sample board: 1/5 lbs
We need to get the weight of the whole square foot.
1/3 square foot is only a part of the whole square foot. Its corresponding weight is 1/5 pounds. We need to divide the area by its corresponding weight to get the total weight per square foot.
1/3 ÷ 1/5
Rules of dividing fractions:
1st fraction: 1/3 ; 2nd fraction: 1/5
1) Get the reciprocal of the 2nd fraction. 1/5 ⇒ 5/1
2) Multiply 1st fraction to the reciprocal of the 2nd fraction
1/3 * 5/1 = 5/3
3) Simplify fraction.
5/3 is an improper fraction. To simplify it, it needs to become a mixed fraction.
5/3 = 1 2/3 pounds.
=
absolute values are always positive.
Answer:
The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.
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 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 mean subtracted by M. So it is 500 - 25.90 = 474.10 milligrams.
The upper end of the interval is the mean added to M. So it is 500 + 25.90 = 525.90 milligrams
The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.
Answer:
P( That it will take over 10 years or more of a year with a rainfall above 50inches) = (0.9938)^10
Step-by-step explanation:
Since the annual rainfall is normally distributed,
Given: that
Mean (µ )= 40
and σ = 4.
Let X be normal random variables of the annual rainfall.
P(that there will be over 10 years or more before a year with a rainfall above 50 inches)
P(>50) = 1-P[X ≤50]
1 - P[X- μ/σ ≤ 50 - 40/4]
=1 - P [Z≤ 5/2]
=1 -Φ(5/2)
=1 - 0.9939
= 0.0062
P( the non occurrence of rainfall above 50 inches)
= 1-0.0062
=0.9938
ASSUMPTION:
P( That it will take over 10 years or more of a year with a rainfall above 50inches) =