D) is the answer i already done that question
Answer:
a) 46.98%
b) increasing rate of change
Step-by-step explanation:
In year 1987, investment worth $29800
In year 1997, investment worth $43800
Rate of change = 1997investment - 1987investment / 1987investment × 100
Rate of change = 43800 - 29800/29800 × 100
= 14000/29800 × 100
= 0.46979 × 100
= 46.98%
Therefore, the rate of change of the investment during the time period is 46.98%.
b) The rate of change of the investment is increasing. This is as a result of the following reasons. First, the value of rate of change is positive. Second, there is no value for rate of change for period before the time period so we cannot compare rate of change.
The statement 'of ' represents multiplication.
thus, baby = 1/40 * 120 = 3 tons.
1 ton is 2000 pounds.
Hence, 3 is 6000 pounds.
Data set: <span>2, 10, 10, 11, 11, 12, 12, 12, 13, 14, 14
</span>
range: 14 - 2 = 12
median: 12
lower quartile = 10
interquartile range = 4
Data set w/o outlier 2: <span>10, 10, 11, 11, 12, 12, 12, 13, 14, 14
range: 14 - 10 = 4
median: 12
lower quartile: 10.5
interquartile range = 3.5
The value that change the most by removing the outlier is A.) THE RANGE.</span>
<span>Each hot dog costs $ 1.50 and each soda costs $ .50. You made a total of $ 78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs and soda's sold. How many hot dogs and soda's were sold?
Quantity Equation: h + s = 87
Value Equation:: 1.5h+0.5s=78.50
Multiply thru the Quantity Equation by 15
Multiply thru the Value Equation by 10.
15h + 15s = 15*87
15h + 5s = 785</span>Subtract and solve for "s":<span>10s = 520
</span>s = 52 (# of soda's sold)
<span>
Solve for h:
h + s = 87
h + 52 = 87
h = 35 (# of hot dog's sold)</span>
