Answer:
The amount that should be in the account after 15 years is $95,321.85
Step-by-step explanation:
According to the given data, we have the following:
monthly amount of $220=R
interest rate is fixed at 2.05%. We require the monthly ineterest rate, hence monthly interest rate= 2.05%/12=0.1708%=0.0017
t=15years×12=180 months
In order to calculate how much should be in the account after 15 years, we would have to use the following formula:
Ap=<u>R(1-(1+i)∧-t)</u>
i
Ap=<u>220(1-(1+0.0017)∧-180)</u>
0.0017
Ap=<u>162,04</u>
0.0017
Ap=$95,321.85
The amount that should be in the account after 15 years is $95,321.85
<u />
<span>-Both box plots show the same interquartile range.
>Interquartile range (IQR) is computed by Q3-Q1.
For Mr. Ishimoto's class, Q3 is 35 and Q1 is 31. 35-31 = 4.
For Ms. Castillo's class, Q3 is 34 and Q1 is 30. 34-30 = 4.
</span><span>-Mr. Ishimoto had the class with the greatest number of students.
>Mr. Ishimoto had 40 students, represented by the last data point of the whiskers.
</span><span>-The smallest class size was 24 students.
>Which was Ms. Castillo's class.</span>
He would be driving approximately 110 km an hour
Answer:
Lance will arrive at the airport at 10:00 AM.
Step-by-step explanation:
Lance left his house at 6:00 AM, and drove to the airport, which is 220 miles away, at 55mph; therefore, the time
it takes Lance to cover the distance of 220 miles is


Thus, it takes Lance 4 hours to reach the airport, and since he departed at 6:00 AM, he will arrive at the airport at (6+4):00 AM + 10:00 AM.
Answer:
a) 0.0228
b) 94.6
Step-by-step explanation:
The formula for calculating a z-score when you are given a random.number of samples is z = (x-μ)/σ/√n
where x is the raw score
μ is the population mean
σ is the population standard deviation
n = random number of samples
Given a normal distribution with u = 75 and o = 40, if you select a sample of n = 16,
a) what is the probability that the sample mean is above 95? (4 d.p.) b)
= x = 95
Hence:
z = 95 - 75/40/√16
= 20/40/4
= 20/10
= 2
Probability value from Z-Table:
P(x<95) = 0.97725
P(x>95) = 1 - P(x<95) = 0.02275
The probability that the sample mean is above 95 to 4 decimal places = 0.0228
b) What is the value, of which there is 97.5% chance that a sample mean is less than that value?
97.5% chance = z score for the confidence interval = 1.96
Hence:
z = (x-μ)/σ/√n
1.96 = x - 75/40/√16
1.96 = x - 75/ 40/4
1.96 = x - 75/10
Cross Multiply
1.96 × 10 = x - 75
19.6 = x - 75
x = 19.6 + 75
x = 94.6
The value, of which there is 97.5% chance that a sample mean is less than that value is 94.6