Answer:
The answer is given below
Step-by-step explanation:
Given that:
The function is an exponential function.
The domain is the set of all independent variables i.e the input values (x values). For an exponential function, the domain is the set of all real numbers. That is:
Domain: x = (-∞, ∞)
The range is the set of all dependent variables i.e the values of y. For an exponential function, the range is the set of all real numbers greater than zero. That is:
Range: y = (0, ∞)
Answer:
She will pay $7,500 interest over the 5 years
She will have to pay back $17,500 in total
Step-by-step explanation:
Let us revise the rule of the simple interest
<em>I</em> = <em>Prt</em> and <em>A</em> = <em>P</em>(1 +<em> rt</em>) , where:
∵ Leah borrows £10,000 over 5 years at a simple interest rate of 15%.
∴ P = 10,000
∴ t = 5
∴ r = 15% = = 0.15
→ Substitute them in the first rule to find the interest
∵ I = 10,000(0.15)(5)
∴ I = 7,500
∴ She will pay $7,500 interest over the 5 years
→ Let us find A
∵ A = 10,000(1 + 0.15×5)
∴ A = 17,500
∴ She will have to pay back $17,500 in total
Answer:
Step-by-step explanation:
Hello!
You have the variable
X: Area that can be painted with a can of spray paint (feet²)
The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²
since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.
a.
P(X>27)
First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)
P(Z>(27-25)/3)
P(Z>0.67)
Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of , you have to do the following conervertion:
P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143
b.
There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².
The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:
X[bar]~N(μ;δ²/n)
So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)
To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.
c.
P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999
d.
No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.
I hope it helps!