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Tamiku [17]
3 years ago
14

A 1/4 mile long suspension bridge is being planned which will require 16 miles of 150 strand (150 wires twisted ) cable to be pl

aced. What is the minimum (ignoring length for twisting) length in km of wire the cable manufacturer needs to have to produce the cable?
Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

out of all of them multimode cables is the best answer

Step-by-step explanation:

You might be interested in
What are numbers that are between 1 and 20 and have exactly 4 factors
vova2212 [387]

Answer:

8 has four factors

Step-by-step explanation:

{1,2,4,8}

1x8=8

2x4=8

Hope this helps!!

3 0
3 years ago
Use pictures and Numbers to show all the ways to take apart 8
Nastasia [14]
2+2+2+2 2×4 8×1 8 4×2
3 0
3 years ago
Noah bought a new car costing $25,350 He took out a 5 year loan at a simple interest rate of 6.5% How much interest did Noah pay
liq [111]

Answer:  8238.75

Work Shown:

i = P*r*t

i = 25350*0.065*5

i = 8238.75

6 0
2 years ago
Read 2 more answers
A wildlife biologist catches, tags, and then releases a number of wasps. Later she catches 55 wasps and finds that 7% are tagged
vesna_86 [32]

Answer:

3.85 tagged wasp

Step-by-step explanation:

6 0
2 years ago
Jenna saves $2,500 per year in an account that earns 10% interest per year, compounded annually. Jenna will have saved in 30 yea
Vsevolod [243]

~~~~~~~~~~~~\textit{Future Value of an ordinary annuity}\\ ~~~~~~~~~~~~(\textit{payments at the end of the period}) \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]

\begin{cases} A= \begin{array}{llll} \textit{accumulated amount}\\ \end{array}\\ pymnt=\textit{periodic payments}\dotfill &2500\\ r=rate\to 10\%\to \frac{10}{100}\dotfill &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &30 \end{cases}

A=2500\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 30}-1}{\frac{0.1}{1}} \right]\implies A=2500\left[ \cfrac{(1.01)^{30}-1}{0.1} \right] \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A\approx 411235.06~\hfill

4 0
2 years ago
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