Answer:
Step-by-step explanation:
Let's first find the exponential function that models the situation in year one. The exponential standard form is
where a is the initial value and b is the growth/decay rate in decimal form. If it is growth it is added to 100% of the initial value; if it is decay it is taken away from 100% of the initial value. We are told that the number of cars in year one was 80 million, so
a = 80 (in millions)
If b is increasing by 10%, then we are adding that amount to the initial 100% we started with to give us 100% + 10% = 110% or, in decimal form, 1.1
The model for our situation is
where y is the number of cars after x years goes by. We want to find the difference between years 3 and 2, so we will use our model twice, replacing x with both a 2 and then a 3 and subtracting.
When x = 2:
and
y = 80(1.21) so
y = 96.8 million cars
When x = 3:
and
y = 80(1.331) so
y = 106.48 million cars
The difference between years 3 and 2 is
106.48 - 96.8 = 9.68 million cars
To combine these terms you would add up the a's and b's.
4a+b+a is the same as 4a+a+b. Anything without a constant in front has a value of 1. You have 4a's and you need to add another a. That equals 5a's. The b is left alone since you can only add like variables.
The answer would be C) 5a +b
Based on the table showing the percentage of households playing games over the net, the average rate of change from 1999 to 2003 is 3.9% per year.
<h3>What is average rate of change?</h3>
This can be found as:
= (27.9 - 12.3) / 4 years
= 3.9% per year
In 2000, the instantaneous rate of change would be:
= (Rate in 2001 - Rate in 1999) / difference in years
= (24.4 - 12.3) / (2001 - 1999)
= 6.05%
Find out more on the average rate of change at brainly.com/question/2263931.
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You have 5 quarters which equals 25 cents each. Which in total comes to $1.25.
You have 15 dimes which is worth 10 cents each.
Which in total comes to $1.50.
$1.50
+$1.25
=$2.75
Answer:
Since the null hypothesis is true, finding the significance is a type I error.
The probability of the year I error = level of significance = 0.05.
so, the number of tests that will be incorrectly found significant is computed as follow: 0.05 * 100 = 5
Therefore, 5 tests will be incorrectly found significant given that the null hypothesis is true.
