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3 years ago

9

There were d passengers in the train. 15% of all of them went in the first car, 20% went in the second car. How many passengers

were in the rest of the train?

1 answer:

vitfil [10]3 years ago

4 0

55% were in the rest of the train. it says "d" though, so I can't tell you the exact number.

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Abby works in the sales department. This month the company is offering a 3-week paid vacation to every employee who sells 230% o

Doss [256]

Multiply 48,000 by 2.3 to get 110,400
She needs to sell $110,400 products next month

8 0

3 years ago

what is the common ratio for the geometric sequence below, written as a fraction? 768, 480, 300, 187.5, …

Crazy boy [7]

Let

$a1=768\\a2=480\\a3=300\\a4=187.5$

Step $1$

Find $\frac{a2}{a1}$

$\frac{a2}{a1} =\frac{480}{768}$

Divide by $96$ numerator and denominator

$\frac{480}{768} =\frac{\frac{480}{96}}{\frac{768}{96}} \\ \\ \frac{480}{768}=\frac{5}{8}$

$a2=a1*\frac{5}{8}$

Step $2$

Find $\frac{a3}{a2}$

$\frac{a3}{a2} =\frac{300}{480}$

Divide by $60$ numerator and denominator

$\frac{300}{480} =\frac{\frac{300}{60}}{\frac{480}{60}} \\ \\ \frac{300}{480}=\frac{5}{8}$

$a3=a2*\frac{5}{8}$

Step $3$

Find $\frac{a4}{a3}$

$a4=187.5 \\ a4=187\frac{1}{2} \\ \\ a4=\frac{375}{2}$

$\frac{a4}{a3} =\frac{\frac{375}{2}}{300} \\ \\ \frac{a4}{a3} =\frac{375}{600}$

Divide by $75$ numerator and denominator

$\frac{375}{600} =\frac{\frac{375}{75}}{\frac{600}{75}} \\ \\ \frac{375}{600}=\frac{5}{8}$

$a4=a3*\frac{5}{8}$

In this problem

The geometric sequence formula is equal to

$a(n+1)=a(n)*\frac{5}{8}\\$

For $n\geq 1$

therefore

the answer is

the common ratio for the geometric sequence above is $\frac{5}{8}$

6 0

4 years ago

Read 2 more answers

if j is 21 inches, k is 12 inches, and angle M measures 62 degrees, then find m using the Law is Cosines. round your answer to t

AURORKA [14]

Answer:

18.7

Step-by-step explanation:

The Law of Cosines is $c^2=a^2+b^2-2ab cos(C)$ , where c is the unknown side length of a triangle, a and b are the remaining two side lengths, and C is the angle opposite to side c. To answer this question just plug in the known values:

$m^2=21^2+12^2-2(21)(12) cos(62)$

Simplify:

$m^2=441+144-504 cos(62)$

$m^2=585-504 cos(62)$

$m=\sqrt{585-504cos(62)}$

m≈18.665...

When rounded to the nearest tenth, m=18.7

6 0

3 years ago

A random sample of 12 recent college graduates reported an average starting salary of $54,000 with a standard deviation of $6,00

Marianna [84]

Answer: a.) $50188 to $57812

Step-by-step explanation: <u>Confidence</u> <u>Interval</u> (CI) is an interval of values in which we are confident the true mean is in.

The interval is calculated as

x ± $z\frac{s}{\sqrt{n} }$

a. For a 95% CI, z-value is 1.96.

Solving:

54,000 ± $1.96.\frac{6000}{\sqrt{12} }$

54,000 ± $1.96\frac{6000}{3.464}$

54,000 ± 1.96*1732.102

54,000 ± 3395

This means the interval is

50605 < μ < 57395

<u>With a 95% confidence interval, the mean starting salary of college graduates is between 50605 and 57395 or </u><u>from 50188 to 57812$.</u>

<u />

b. The mean starting salary for college students in 2017 is $50,516, which is in the confidence interval. Therefore, since we 95% sure the real mean is between 50188 and 57812, there was no significant change since 2017.

4 0

3 years ago

Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.

Eduardwww [97]

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

8 0

3 years ago