The correct format of the question is
At the end of 2006, the population of Riverside was 400 people. The population for this small town can be modeled by the equation below, where t represents the number of years since the end of 2006 and P represents the number of people.
Based on this model, approximately what was the increase in the population of Riverside at the end of 2009 compared to the end of 2006?
(A) 291
(B) 691
(C) 1040
(D) 1440
Answer:
The increase in the population at the end of 2009 is 291 people
Step-by-step explanation:
We are given the equation as
where
P = No of People
t= No of Years
it is given that in the year 2006 the population is 400
this will only happen when we take t= 0
so for
Year value of t
2006- t = 0
2007- t = 1
2008- t = 2
2009 t = 3
No of people in 2009 will be

= 400*1.728
P = 691.2
Since the equation represents no of people so it can't be in decimals, Therefore the population will be 691
Increase = P(2009) - P(2006)
= 691 - 400
= 291
The increase in the population at the end of 2009 is 291 people.
When given the scale factor and the factored size of the object, to find the original size, divide the factored size by the scale factor:
0.8 / 3.1 = 0.258 inches
Answer:
There is no sufficient evidence to reject the company's claim at the significance level of 0.05
Step-by-step explanation:
Let
be the true mean weight per apple the company ship. We want to test the next hypothesis
vs
(two-tailed test).
Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by
which is normally distributed. The observed value is
. The rejection region for
is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.
Answer:
13, 14
Step-by-step explanation:
The parameters of the numbers are;
A whole number value = 2 × Another number + 6
The sum of the two numbers is less than 50
Given that the first number is equal to more than twice the second number, we have that the first number is the larger number, while the second number is the smaller number
Where 'x' represents the second number, we get;
x + 2·x + 6 < 50
Simplifying gives;
3·x + 6 < 50
x < (50 - 6)/3 = 14.
x < 14.
Therefore, the numbers for which the inequality holds true are numbers less than 14.
. From the given option, the numbers are 13, and 14.
Answer: 1/6
Step-by-step explanation: