Answers

MN

Answered

Susan Phillips made two $1200 deposits every year (i.e., semiannual) for 10 years. If the investment pays a return of 6% compounded semiannually, how much interest would Susan's investment earn during the 10 years? Use Tables 23-1A and 23-1B or a calculator.​

On Jun 29, 2024


$1,200 × 26.87037 = $32,244.44; $1,200  20 = $24,000;
$32,244.44 -$24,000 = $8,244.44 interest earned
MN

Answered

Cameron Insurance Company insured Driver Gifford at an annual premium of $740. After 6 months, Gifford sold the car and canceled the insurance. Cameron Insurance Company refunded the remaining half of the premium at the short rate based on a penalty of 15%. Compute the amount of the short-rate refund.

On Jun 10, 2024


$259
MN

Answered

Evaluate to six-figure accuracy:
1.056\sqrt[6]{1.05}61.05

On May 17, 2024


1.00816
MN

Answered

Use the properties of logarithms to expand ln⁡2xx−6\ln \sqrt { \frac { 2 x } { x - 6 } }lnx62x .

A) (ln⁡2+ln⁡x−ln⁡(x−6) ) 1/2( \ln 2 + \ln x - \ln ( x - 6 ) ) ^ { 1 / 2 }(ln2+lnxln(x6) ) 1/2
B) (ln⁡2⋅ln⁡xln⁡x−ln⁡6) 1/2\left( \frac { \ln 2 \cdot \ln x } { \ln x - \ln 6 } \right) ^ { 1 / 2 }(lnxln6ln2lnx) 1/2
C) 12(ln⁡2+ln⁡x−ln⁡xln⁡6) \frac { 1 } { 2 } \left( \ln 2 + \ln x - \frac { \ln x } { \ln 6 } \right) 21(ln2+lnxln6lnx)
D) 12(ln⁡2+ln⁡x−ln⁡(x−6) ) \frac { 1 } { 2 } ( \ln 2 + \ln x - \ln ( x - 6 ) ) 21(ln2+lnxln(x6) )
E) 12ln⁡2+ln⁡x−ln⁡(x−6) \frac { 1 } { 2 } \ln 2 + \ln x - \ln ( x - 6 ) 21ln2+lnxln(x6)

On May 14, 2024


D
MN

Answered

In a Wilcoxon rank sum test,the two sample sizes are 4 and 6,and the value of the Wilcoxon test statistic is T = 20.If the test is a two-tail and the level of significance is α = 0.05,then:

A) the null hypothesis will be rejected.
B) the null hypothesis will not be rejected.
C) the alternative hypothesis will not be rejected.
D) not enough information has been given to answer this question.

On May 10, 2024


B