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Veseljchak [2.6K]

2 years ago

10

A railroad crew can lay 7 miles of track each day. They need to lay 203 miles of track. The length, L (in miles), that is left t

o lay after d days is given by the
following function.
L(d) = 203 - 7d
(How many days will it take the crew to lay all the track?)
(How many miles of track does the crew have left to lay after 18 days?)

1 answer:

swat322 years ago

7 0

Answer:

It will take 29 days to lay the track

After 18 days the crew still has 77 miles of track left to lay

Step-by-step explanation:

203÷7

=29 days

L= 203 - 7(18)

L= 203 - 126

L= 77

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<h2>Answer</h2>

0.43

<h2>Explanation</h2>

Remember that $RelativeFrecuency=\frac{Frecuency}{SumOfAllFrecuencies}$

Since the problem is telling us "Among tenth graders", we must focus on the 10th graders row only. From the row, we can infer that the frequency is the number of 10th graders who prefer going to sporting events, so $Frecuency=6$ . Now, the sum of all frequencies will be the sum of all the 10th graders, so $SumOfAllFrecuencies=6+8=14$ . Let's replace the values:

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2 years ago

Seventy cards are numbered 1 through 70, one number per card. One card is randomly selected from the deck. What is the probabili

miskamm [114]

Given:

Seventy cards are numbered 1 through 70, one number per card. One card is randomly selected from the deck.

To find:

The probability that the number drawn is a multiple of 3 and a multiple of 5.

Solution:

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Multiple of 3 from 1 to 70 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69.

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2 years ago

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Answer:

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So x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

Substituting the values of the coordinates, we have

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3 0

1 year ago

If 6j-5k =11 and 5j-6k =22, then what is the value of 2j+2k??

dsp73

Answer:

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Step-by-step explanation:

6j-5k=11

5j-6k=22

-----------------

5(6j-5k)=5(11)

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-----------------------

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-----------------------

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------------------

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2 years ago

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Answer:

$x=\pm\sqrt{3}$ and they are actual solutions

Step-by-step explanation:

we have

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Factor the denominators both sides

$\frac{x^2}{2(x-3)}=\frac{9}{6(x-3)}$

Simplify

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<u><em>Verify</em></u>

1) For $x=\sqrt{3}$

$\frac{\sqrt{3}^2}{2(\sqrt{3}-3)}=\frac{9}{6(\sqrt{3}-3)}$

$\frac{3}{2(\sqrt{3}-3)}=\frac{9}{6(\sqrt{3}-3)}$

$18=18$ ---> is true

therefore

$x=\sqrt{3}$ ----> is an actual solution

2) For $x=-\sqrt{3}$

$\frac{-\sqrt{3}^2}{2(-\sqrt{3}-3)}=\frac{9}{6(-\sqrt{3}-3)}$

$\frac{3}{2(-\sqrt{3}-3)}=\frac{9}{6(-\sqrt{3}-3)}$

$18=18$ ---> is true

therefore

$x=-\sqrt{3}$ ----> is an actual solution

therefore

3 0

2 years ago

Read 2 more answers