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kramer
3 years ago
10

Please help fast will mark 5 stars

Mathematics
2 answers:
harkovskaia [24]3 years ago
8 0

Answer:

1 solution

Step-by-step explanation:

The intersection is the one solution.

vredina [299]3 years ago
5 0

Answer:

1 solution

Step-by-step explanation:

The lines intersect 1 time

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X=16-4y. systems of equations 3x+4y=8<br> substitution
alina1380 [7]

Answer:

x=-4  y=5

Step-by-step explanation:

Rewrite equations:

x=−4y+16;3x+4y=8

Step: Solvex=−4y+16for x:

x=−4y+16

Step: Substitute−4y+16forxin3x+4y=8:

3x+4y=8

3(−4y+16)+4y=8

−8y+48=8(Simplify both sides of the equation)

−8y+48+−48=8+−48(Add -48 to both sides)

−8y=−40

−8y

−8

=

−40

−8

(Divide both sides by -8)

y=5

Step: Substitute5foryinx=−4y+16:

x=−4y+16

x=(−4)(5)+16

x=−4(Simplify both sides of the equation)

Answer:

x=−4 and y=5

6 0
2 years ago
two cars leave Memphis from exactly the same spot. one traveling north at 70 mph and the other traveling south at 65 mph. how lo
Gemiola [76]

Answer:

2 hours

Step-by-step explanation:

270 = (70+65) * t

270= 135t

t = 2

8 0
3 years ago
An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0
3 years ago
Jill wants to find the quotient use multiplication and distributive property to help Jill find the quotient of 144 divided by 8
UkoKoshka [18]
120+24 divided by 8=
144 divided by 8= 18 because 18 times 8 equals 144
5 0
3 years ago
Read 2 more answers
2/20 divided by 8/10
raketka [301]

Answer:

1/8

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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