question 15 (4 points) a new car is worth $60 000. after 5 years, the car was worth $35 429.40. calculate…

Question 15

Explanation:

Step1: Set up depreciation formula

The formula for depreciation is $A = P(1 - r)^t$, where $A$ is the final value, $P$ is the initial value, $r$ is the annual - depreciation rate, and $t$ is the number of years. We know that $P=$60000$, $A = $35429.40$, and $t = 5$. So, $35429.40=60000(1 - r)^5$.

Step2: Isolate $(1 - r)^5$

Divide both sides of the equation by 60000: $\frac{35429.40}{60000}=(1 - r)^5$. Calculate $\frac{35429.40}{60000}=0.59049=(1 - r)^5$.

Step3: Solve for $(1 - r)$

Take the fifth - root of both sides: $1 - r=\sqrt[5]{0.59049}$. Using a calculator, $\sqrt[5]{0.59049}\approx0.89999\approx0.9$.

Step4: Solve for $r$

Rearrange the equation $1 - r = 0.9$ to get $r=1 - 0.9 = 0.1$ or $10%$.

Answer:

$10%$

Question 16

Explanation:

Step1: Recall pH formula

The formula for pH is $pH=-\log[H^+]$, where $[H^+]$ is the hydrogen - ion concentration. Given $[H^+]=0.005\ mol/L$.

Step2: Calculate pH

Substitute $[H^+]=0.005$ into the formula: $pH =-\log(0.005)$. Using a calculator, $\log(0.005)=\log(5\times10^{-3})=\log5+\log(10^{-3})\approx0.69897 - 3=-2.30103$. Then $pH=-(-2.30103)=2.30103$.

Step3: Round to one decimal place

Rounding $2.30103$ to one decimal place gives $2.3$.

Answer:

$2.3$