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3 years ago

There are 60 seats on a train. 6 of the seats are empty. What percentage of the seats on the train are empty?

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

6 0

Answer:

10%

Step-by-step explanation:

Set up a proportion: $\frac{6}{60} = \frac{x}{100}$
Cross multiply, then divide: 6 × 100 = 600, 600 ÷ 60 = 10
10/100 = 10%

I hope this helps!

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The value of the expression is

elixir [45]

Answer:

P: parentheses (12 - 8) = 4

E: exponents (3^2) = 9

M: multiplication (12-8)/2 * 4 = 8

D: division (12-8)/2 = 2

A: addition (7 + 9 + 8) = 24

S: subtraction (blank)

Following the order of operations, our final answer is: 24

3 0

3 years ago

Read 2 more answers

Which description does NOT guarantee that a quadrilateral is a square?

Ivahew [28]

Let's go through the choices one by one

------------------------------------------
Choice A

If all sides are congruent, then this figure is a rhombus (by definition). If all angles are congruent, then we have a rectangle. Combine the properties of a rhombus with the properties of a rectangle and we have a square.

In terms of "algebra", you can think
rhombus+rectangle = square

Or you can draw out a venn diagram. One circle represents the set of all rhombuses; another circle represents the set of all rectangles. The overlapping region is the set of all squares. The overlapping region is inside both circles at the same time.

So we can rule out choice A. This guarantees we have a square when we want something that isn't a guarantee.

------------------------------------------
Choice B

If we had a parallelogram with perpendicular diagonals, then we can prove that we have a rhombus (all four sides congruent). However, we don't know anything about the four angles of this parallelogram. Are they congruent? We don't know. So we can't prove this figure is a rectangle. The best we can say is that it's a rhombus. It may or may not be a rectangle. There isn't enough info about the rectangle & square part.

This is why choice B is the answer. We have some info, but not enough to be guaranteed everytime.

------------------------------------------
Choice C

This is a repeat of choice A. Having "all right angles" is the same as saying "all angles congruent". This is because "right angle" is the same as saying "90 degrees". So we can rule out choice C for identical reasons as we did with choice A.

------------------------------------------
Choice D

As mentioned before in choice A, if we know that a quadrilateral is a rectangle and a rhombus at the same time, then the figure is also a square. This is always true, so we are guaranteed to have a square. We can cross choice D off the list.

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

Once again, the final answer is choice B

3 0

4 years ago

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

igor_vitrenko [27]

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

$m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7$

So, the solution of the equation is

$y = Ae^{m_{1} t} + Be^{m_{2} t}$

where m₁ = 3 and m₂ = -7.

So,

$y = Ae^{3t} + Be^{-7t}$

Also,

$y' = 3Ae^{3t} - 7e^{-7t}$

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

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4 years ago

There are 34 animals in a farmyard. some are cows and some are chickens. in total there are 104 legs. how many chickens are ther

Korvikt [17]

There are 18 cows and 16 chickens.

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3 years ago

George wants to know if two pieces of wire are the same length. one wire is 6/8 yard. the other is 3/4 yard. draw a number Line

wlad13 [49]

They are the same length. You need to find the common denominator (8) and multiply both the numerator and denominator on 3/4 to ger 6/8. Than, just draw a numberline comparing the lengths! Hope this helped

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4 years ago