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3 years ago

If there were 35,000 fatal traffic accidents, how many would be expected to involve drivers in the 55-64 age group?

Mathematics

1 answer:

olasank [31]3 years ago

5 0

You need more information to answer this question. Is there a table or percentage for that age group?

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<img src="https://tex.z-dn.net/?f=Simplify%3A%20%5Cfrac%7B%205%C3%97%2825%29%5E%7Bn%2B1%7D%20-%2025%20%C3%97%20%285%29%5E%7B2n%7

Katen [24]

$\green{\large\underline{\sf{Solution-}}}$

<u>Given expression is </u>

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

<u>Hence, </u>

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

4 0

2 years ago

1 pt) find the average rate of change of the function over the given intervals. h(t)=\cot (t);

blondinia [14]

$\text{Consider the function}\\
\\
h(t)=\cot(t) \text{ on the interval }\left [ \frac{\pi}{4}, \ \frac{3\pi}{4} \right ]\\
\\
\text{we know that the average rate of change of a funtion f(x) over the }\\
\text{interval [a, b] is given by}\\
\\
f_{avg}=\frac{f(b)-f(a)}{b-a}\\
\\
\text{so using this, the average rate of change of the given function is}$

$h_{avg}=\frac{h\left ( \frac{3\pi}{4} \right )-h\left ( \frac{\pi}{4} \right )}{\frac{3\pi}{4}-\frac{\pi}{4}}\\
\\
=\frac{\cot \left ( \frac{3\pi}{4} \right )-\cot \left ( \frac{\pi}{4} \right )}{\frac{3\pi-\pi}{4}}\\
\\
=\frac{(-1)-(1)}{\frac{2\pi}{4}}\\
\\
=\frac{-2}{\frac{\pi}{2}}\\
\\
\text{Average rate of change of function}=\frac{-4}{\pi}$

6 0

3 years ago

A student charges twenty dollars per hour to type college papers. The student's average typing speed is fifty words per minute.

coldgirl [10]

Answer:

The student should charge $210 for typing the paper ⇒ answer C

Step-by-step explanation:

* Lets explain how to solve the problem

- The rate of the student's typing is 50 words per minute

- The students charges twenty dollars per hour to type the college

papers

- There is a paper contains 31,500 words

- We want to now how much the student should charge for it

* At first lets find his rate per hour

∵ 1 hour = 60 minutes

∴ 1 minute = 1/60 hours

∵ His rate = 50 words / 1 minute

- Lets change the minute to hour

∴ His rate = 50/(1 × 1/60) = 3000 words per hour

* Lets find the time he needs to finish the paper

∵ The paper contains 31,500 words

∵ Time = The number of words ÷ his rate of typing

∵ his rate is 3000 words per hour

∴ Time = 31,500 ÷ 3000 = 10.5 hours

∴ He needs 10.5 hours to finish the paper

∵ He charges $20 per hour

∵ The time to finish the paper is 10.5 hours

∵ The price of typing the paper = the charge per hour × number

of hours

∴ The price of typing the paper = 20 × 10.5 = 210

∴ The student should charge $210 for typing the paper

7 0

3 years ago

Read 2 more answers

5. Trenton is going to close and wrap the box

LenKa [72]

Answer:

Answer in the picture posted

5 0

2 years ago

Please help ASAP! !!!!!!!!!

vredina [299]

= (8 x 2) x (10^7 x 10^6)
= 16 x 10^13
= 1.6 x 10^14

answer is C
1.6 x 10^14

8 0

3 years ago