The answer is $1,026.
Solving: Use Simple interest formula, I=Prt.
Make the rate into a decimal (1.6% -> 0.016)
plug in the numbers into the formula. 950 x 0.016 x 5 = 76$.
76$ will be our interest, now to find the balance we would have to add the the starting money (950) with our interest (76$), resulting in our answer of $1,026 after 5 years.
Answer:
Step-by-step explanation:
You have to divide the TOTAL underweight which is 130, into the total of individuals which is 1000. 130/1000= 0.13 or 13.0%
Answer C.
I just took this quiz today
Answer:
There is no sufficient evidence to support the claim.
Step-by-step explanation:
Given the data:
7.91, 7.85, 6.82, 8.01, 7.46, 6.95, 7.05, 7.35, 7.25, 7.42
Sample size, n = 10
The sample mean, xbar = ΣX/ n = 74.07 / 10 = 7.407
The sample standard deviation, s = 0.41158 ( from calculator)
The hypothesis :
H0 : μ = 7
H0 : μ ≠ 7
The test statistic :
(xbar - μ) ÷ (s/√(n))
(7.047 - 7) ÷ (0.41158/√(10))
0.047 / 0.1301530
Test statistic = 0.361
Testing the hypothesis at α = 0.05
The Pvalue ;
df = n - 1 ; 10 - 1 = 9
Two tailed test
Pvalue(0.361, 9) = 0.7263
Since the Pvalue > α ; we fail to reject the Null and conclude that there isn't sufficient evidence to support the claim.
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
