Answer:
The annual rate of interest is 2.56%
Step-by-step explanation:
Given as :
Principal Amount investing every week = $25
Principal Amount investing in 5 years = $25 × 5 × 52 = p = $6500
The Amount adds after 5 years = A = $7380
Time period = t = 5 years
Let annual rate of interest = r %
<u>Now, According to question</u>
Amount after 5 years = Principal ×
Or, A = p ×
Or, $7380 = $6500 ×
Or, =
Or, 1.135 =
Or, = 1 +
Or, 1.0256 = 1 +
Or, 1.0256 - 1 =
Or, 0.0256 =
∴ r = 0.0256 × 100
i.e r = 2.56
So, The rate of interest = r = 2.56
Hence, The annual rate of interest is 2.56% Answer
Answer: The variance of the data is 0.25, therefore second option is correct.
Explanation:
From the given graph it is noticed that the graph is symmetric along x = 4 or it is a bell shaped graph. So, the data follows the normal distribution.
The graph of normal distribution is bell shaped graph symmetric along the mean. The graph of normal distribution is shown below.
Since the given graph is symmetric along x = 4 therefore the mean of the data is 4.
Let the variance be . So, the given data follows normal distribution with mean 4 and variance .
In the below graph the point immediate left to the mean is and in the given graph it 3.5
Therefore, the variance of the data is 0.25, therefore second option is correct.
Using a system of equations, the expression or the number of touchdowns he scored in his second year is: 2(t - 1)
<h3>What is a system of equations?</h3>
A system of equations is when two or more variables are related, and equations are built to find the values of each variable.
In his first year, he scored t touchdowns. In the second, the amount was two less than twice the number of touchdowns he scored in his rookie year, hence the expression is:
2t - 2 = 2(t - 1)
More can be learned about a system of equations at brainly.com/question/24342899
Answer:
See verification below
Step-by-step explanation:
We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that
Now, differentiate to obtain
To obtain the required form, extract a factor in both the numerator and denominator: