Answer:
$20,086.35
Step-by-step explanation:
To calculate the maturity value by compound interest, we will use the formula
where,
A = Maturity amount
P = Principal amount = $10,000
r = rate of interest = 4.65% = 0.0465
n = number of compounding periods = 365
t = time in years = 15 years
Now substituting the values,
=
= 10,000(2.008635)
= 20086.353758 ≈ $20,086.35
The final value of your investment would be $20,086.35.
Answer:
Step-by-step explanation:
Mark the two points (-1,7) and (1,-1) on the graph. Then draw a straight line between them. To determine the equation that goes through these two points, we can use the two given points to find the slope of the line. The standard form of a straight line equation is
y = mx + b,
where m is the slope and y is the y-intercept (the value of y when x = 0).
Slope is also known as the "Rise"/"Run" - the change in y divided by the change in x. We can use the two points to calculate this:
Rise (-1-(7) = -8 Run = (1 - (-1) = 2
The slope is therefore (-8/2) or -4.
y = -4x + b
We can find b by entering either of the two points in y = -4x + b and solve for b. I'll use (1,-1) since I have my 1's multiplication table memorized
y = -4x + b
-1 = -4(1) + b
b = 3
The straight line equation that connects the two points is
y = -4x + 3
You can graph this equation (e.g., on DESMOS) to see how it intersects the points. <u>[Attached]</u>
The coordinates of the y intercept are (0,3).
Answer:
d) The difference exists due to chance since the test statistic is small
Step-by-step explanation:
From the given information:
Population mean = 178 cm
the sample mean = 177.5 cm
the standard deviation = 2
the sample size = 25
The null hypothesis and the alternative hypothesis can be computed as:
Null hypothesis:
Alternative hypothesis:
The t-test statistics is determined by using the formula:
Degree of freedom df = n- 1
Degree of freedom df = 25 - 1
Degree of freedom df = 24
At the level of significance ∝ = 0.05, the critical value = 2.064
Decision rule: To reject the null hypothesis if the test statistics is greater than the critical value at 0.05 level of significance
Conclusion: We fail to reject the null hypothesis since the test statistics is lesser than the critical value and we conclude that the difference exists due to chance since the test statistic is small