2. ½ of a group of chickens tried to cross the road. Only ¾ of those chickens made it to the other side. What fraction of the or
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Isolating 2abCos(c) on one side of the equation and using the given values of a, b and c we can find the answer to this question as shown below:
Answer:
<em>Mrs. Adams will earn $3,120 of interest at the end of year 8.</em>
Step-by-step explanation:
<u>Simple Interest</u>
In simple interest, the money earns interest at a fixed rate, assuming no new money is coming in or out of the account.
We can calculate the interests earned by an investment of value A in a period of time t, at an interest rate r with the formula:
Mrs. Adams deposited an amount of A=$12,000 into an account that earns an annual simple interest rate of r=3.25%. We must find the interest earned in t=8 years. The interest rate is converted to decimal as:
The interest is then calculated:
Mrs. Adams will earn $3,120 of interest at the end of year 8.
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01
X[bar] ±
174.5 ±
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!