Answer:
Samantha has 8 quarts of milk. The ratio of quarts to gallons is 4:1. How many gallons of milk does Samantha have? x a. 1 over 32 x b. begin mathsize 16px style 1 half end style x c. 2 x d. 32
Step-by-step explanation:Samantha has 8 quarts of milk. The ratio of quarts to gallons is 4:1. How many gallons of milk does Samantha have? x a. 1 over 32 x b. begin mathsize 16px style 1 half end style x c. 2 x d. 32Samantha has 8 quarts of milk. The ratio of quarts to gallons is 4:1. How many gallons of milk does Samantha have? x a. 1 over 32 x b. begin mathsize 16px style 1 half end style x c. 2 x d. 32Samantha has 8 quarts of milk. The ratio of quarts to gallons is 4:1. How many gallons of milk does Samantha have? x a. 1 over 32 x b. begin mathsize 16px style 1 half end style x c. 2 x d. 32
Make each fraction have a like denominator
7/9 * 8/8= 56/ 72
7/8 * 9/9 = 63/72
18 + 14 + (56+63)/72
32 + 119/72
32 + 1 + 47/72
33 and 47/72
Answer:
75% of the data will reside in the range 23000 to 28400.
Step-by-step explanation :
To find the range of values :
We need to find the values that deviate from the mean. Since we want at least 75% of the data to reside between the range therefore we have,
Solving this, we would get k = 2 which shows the value one needs to find lies outside the range.
Range is given by : mean +/- (z score) × (value of a standard deviation)
⇒ Range : 25700 +/- 2 × 1350
⇒ Range : (25700 - 2700) to (25700 + 2700)
Hence, 75% of the data will reside in the range 23000 to 28400.
(4,5),(-3,-1)
slope = (-1 - 5) / (-3 - 4) = -6/-7 = 6/7
y - y1 = m(x - x1)
slope(m) = 6/7
using points (-3,-1)...x1 = -3 and y1 = -1
now we sub...pay attention to ur signs
y - (-1) = 6/7(x - (-3) =
y + 1 = 6/7(x + 3) <===
<h3>
Answer:</h3>
- B. f(x) = 3,000(0.85)^x
- $1566.02
<h3>
Step-by-step explanation:</h3>
Part A
At the end of the year, the value of the computer system is ...
... (beginning value) - 15% · (beginning value) = (beginning value) · (1 - 0.15)
... = 0.85 · (beginning value)
Since the same is true for the next year and the next, the multiplier after x years will be 0.85^x. Then the value after x years is ...
... f(x) = (beginning value) · 0.85^x
The beginning value is given as $3000, so this is ...
... f(x) = 3000·0.85^x
____
Part B
For x=4, this is ...
... f(4) = 3000·0.85^4 = 3000·0.52200625 ≈ 1566.02
The value after 4 years is $1566.02.
