D=5 evaluate 4d <br> Help !!!

Ninety-three passengers rode in a train from City A to City B. Tickets for regular coach seats cost $115. Tickets for sleeper ca

masya89 [10]

To solve this, we must set up a Systems of Equations problem
Let's say C stands for Coach seats bought and S stands for sleeper cars bought
Here are our equations we have enough information to make.
c + s = 93
115c + 290s = $20,320
Lets solve for Coach seats first
To do that we need a value for s
In the first equation, s = 93 - c
Lets plug that in for S in the second equation
115c + 290(93 - c) = $20,320
Distribute and simplify
115c + 26,970 - 290c = $20,320
Combine like terms
-175c + 26,970 = $20,320
-175c = -6650
divide
c = $\frac{-6650}{-175}$
c = 38
We just found out 38 coach seats were bought.
Plug in 38 for C in this equation ⇒ 115c + 290s = $20,320 to find S
115(38) + 290s = $20,320
Distribute and Simplify
4,370 + 290s = $20,320
290s = $15,950
Divide
S = $\frac{15950}{290}$
S = 55
We just completed the problem!
38 Coach seats were bought and
55 Sleeper car seats were bought.

5 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago

-2.8f+0.4f-11-4=? <br><br><br><br> HELP

Alexxandr [17]

<h2>Answer: −2.4f − 15</h2><h2>_____________________________________</h2><h3>Honey, just simplify the expression to get your answer.</h3><h3>−2.4f − 15</h3><h2>_____________________________________</h2>

Hope you have a good day, Loves!~ <3

6 0

3 years ago

Plz help meeeee plz

Sauron [17]

Answer:

Less than

Step-by-step explanation:

Its less than because the angle of LS is 98 while MS is 95, so LS is longer

3 0

3 years ago

The area of a triangle is Half the area of a rectangle with the same base and height *

Vaselesa [24]

Answer:

Always.

Step-by-step explanation:

That is the reason why when you are trying to calculate the area of a triangle you do the base x height divided by two or in half. If you printed out the triangle twice then cut it out and put the two together it would be a rectangle.

5 0

3 years ago

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D=5 evaluate 4d Help !!!