Answer:
$30.88
Step-by-step explanation:
The account value is given by ...
A = P(1 +r/n)^(nt)
where P is the principal invested, r is the annual interest rate, t is the number of years, n is the number of times interest is compounded per year.
The amount of interest earned is the account value less the initial investment:
I = A - P = P(1 +r/n)^(nt) -P = P((1 +r/n)^(nt) -1)
Filling in the given values, we get ...
I = 500((1 +.03/12)^(12·2) -1) = 500(1.0025^24 -1) ≈ $30.88
The amount of interest earned is $30.88.
How many times a particular number is a zero<span> for a given polynomial. For example, in the polynomial function f(x)=(x–3)4(x–5)(x–8)2, the </span>zero<span> 3 has </span>multiplicity<span> 4, 5 has </span>multiplicity<span> 1, and 8 has </span>multiplicity<span> 2. Although this polynomial has only three </span>zeros<span>, we say that it has seven </span>zeros<span> counting </span><span>multiplicity</span>
Answer:
36
Step-by-step explanation:
3*4^(2) - (4*18)/(6)
Answer:
the one selected, the one on top
Step-by-step explanation:
Answer:
Step-by-step explanation:
Proportion of retired people under the age of 65 would return to work on a full-time basis if a suitable job were available = 60/100 = 0.6 = P
Null hypothesis: P ≤ 0.6
Alternative: P > 0.6
First, to calculate the hypothesis test, lets workout the standard deviation
SD = √[ P x ( 1 - P ) / n ]
where P = 0.6, 1 - P = 0.4, n = 500
SD = √[ (0.6 x 0.4) / 500]
SD = √ (0.24 / 500)
SD = √0.00048
SD = 0.022
To calculate for the test statistic, we have:
z = (p - P) / σ where p = 315/500 = 0.63, P = 0.6, σ = 0.022
z = (0.63 - 0.6) / 0.022
z = 0.03/0.022
z = 1.36
At the 2% level of significance, the p value is less than 98% confidence level, thus we reject the null hypothesis and conclude that more than 60% would return to work.
