Answer:
Step-by-step explanation:
A)
"In the last 2.5 years, the water level has decreased 0.75 feet."
In the very first statement it has been stated that over the last 2.5 years, the water level in a lake has changed by -0.3 feet per year. That means in the last 2.5 years , total decrease in level is as follows
2.5 x - 0.3 = - .75 ft .
So the statement is true.
B)
"The overall change of the water level after 7.5 years will be -2.75 feet."
This statement appears to be false based on information provided
No prediction regarding rate of decrease for a period beyond 5 year has been made so any statement with respect to a timeline of 7.5 years may turn out to be false.
C)
"The change in water level each year for the next 5 years will be -0.4 feet.
"
This statement prima-facie appears to be true because rate of decrease stated in the statement is 0.4 ft /year for next 5 years .
D )
"The overall change in water level for the next 5 years is equal to 5(0.4)"
this statement is true
as change in one year = .4
change in 5 years = 5 x .4
The length of the fifth piece is 2 3/4 inches.
Given:
1st piece of wood = 1 1/2 inches
2nd piece of wood = 1 13/16 inches
3rd piece of wood = 2 1/8 inches
f(n) = 1st term + common difference(n-1)
To get the common difference, we need to deduct the fraction from its preceding fraction.
1st piece: 1 1/2 = (2*1 + 1)/2 = 3/2
2nd piece: 1 13/16 = (16*1+13)/16 = 29/16
3rd piece: 2 1/8 = (8*2+1)/8 = 17/8
29/16 - 3/2 = 29/16 - (3/2 * 8/8) = 29/16 - 24/16 = (29-24)/16 = 5/16
17/8 - 29/16 = (17/8 *2/2) - 29/16 = 34/16 - 29/16 = (34-29)/16 = 5/16
Find: 5th piece's measurement
1st term : 1 1/2 or 3/2
common difference : 5/16
f(5) = 3/2 + 5/16 (5-1)
= 3/2 + 5/16 (4)
= 3/2 + (5*4)/16
= 3/2 + 20/16
= (3/2 * 8/8) + 20/16
= 24/16 + 20/16
= (24 + 20)/16
f(5) = 44/16
f(5) = 2 12/16 simplified to 2 3/4 inches.
Given that the scale readings of the laboratory scale is normally distributed with an unknown mean and a standard deviation of 0.0002 grams.
Part A:
If the weight is weighted 5 times and the mean weight is 10.0023 grams, the 98% confidence interval for μ, the true mean of the scale readings is given by:
Thus, we are 98% confidence that the true mean of the scale readings is between 10.0021 and 10.0025.
Part B:
To get a margin of error of +/- 0.0001 with a 98% confidence, then
Therefore, the number of <span>measurements that must be averaged to get a margin of error of ±0.0001 with 98% confidence is 22.</span>
Answer:
1 and 1/16 gallons
Step-by-step explanation:
