Answer:
503.525 gallon ( approx )
Step-by-step explanation:
Given,
Water hold by of the water tank = 5 gallon
We know that, 1 gallon = 133.228 ounces
⇒ Water hold by of tank in ounces = 5 × 133.228 = 666.139 ounces,
Here, it is given that, 6 bottles that each hold 16 ounces and a pitcher that holds 1/2 gallon is used from tank,
Since, the quantity of water hold by6 bottle in which each hold 16 ounces = 6 × 16 = 96 ounces,
Also, 1/2 gallon = 1/2 × 1 gallon = 1/2 × 133.228 = 66.614 ounces
Thus, the total consumed water = 96 + 66.614 = 162.614 gallon
Hence, the quantity of water left in the tank = Water hold by of tank in ounces - the total consumed water
= 666.139 - 162.614
= 503.525 gallon
B. (8, 1), (-5, 4), (2, 1), (8, 2)
C. (2,4), (-5, 2), (8, 1), (-6, 2)
D.
(2,0).(-5, 3), (8, 1), (-5,5)
Answer: OPTION C
Step-by-step explanation:
A relation is a function when each input value has one and only output value.
For option A:
(2,4), (-5, 6), (2, 3), (-6, 2)
You can observe that the input value "2" has two output values. Then it is not a function.
For option B:
(8, 1), (-5, 4), (2, 1), (8, 2)
You can observe that the input value "8" has two output values. Then it is not a function.
For option C:
(2,4), (-5, 2), (8, 1), (-6, 2)
You can observe that each input value has only one output values. Then it is a function.
For option D:
(2,0).(-5, 3), (8, 1), (-5,5)
You can observe that the input value -5 has two output values. Then it is not a function.
For a set of data to be a function each x value has to have different y values.
A. (2,4), (-5, 6), (2, 3), (-6, 2)
B. (8, 1), (-5, 4), (2, 1), (8, 2)
D.(2,0).(-5, 3), (8, 1), (-5,5)
The bolded coordinates are the ones of each data set that makes it not a function. As you can see the x values all have different y values.
This leaves C. as the answer!
Hope this helped!
O a hypothesis
O descriptive analysis
O univariate analysis
Question 10
A.
–8.04
B.
–2.5
C.
–0.7
D.
3.16
Answer:
range=u ± 3.09 sd
Step-by-step explanation:
Given:
mean, u= 26.8 mpg
standard deviation, sd=72 mpg
% contained in interval = 99.8%
the interval for 99.8% of the values of a normal distribution is given by
mean ± 3.09 standard deviation= u ± 3.09 sd
=26.8 ± 3.09(72)
=26.8 ± 222.48
= 249.28 , -195.68
range=u ± 3.09 sd = 249.28 , -195.68 !