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3 years ago

What property is 1/5+1/8=1/8+1/5?

1 answer:

6 0

Answer: This is the <u>Communitve Property</u> because the equation has the same thing on both sides, but the numbers "flip flopped" or switched around.

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If h(x) = 5 x and k(x)=1/x, which expression is equivalent to (koh)(x)?

Lyrx [107]

May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x

1) If h(x) = 5x and k(x) = 1/x

Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)

2) If h(x) = 5 + x and k (x) = 1/x

Then (k o h)(x) =k ( h(x) ) = k (5+x) = 1 / [5 + x]

4 0

3 years ago

Read 2 more answers

Question 1

Galina-37 [17]

Answer:

Did you find the answer???

5 0

2 years ago

Harriet earns the same amount of money each day. Her gross pay at the end of 7 work days is 35h + 56 dollars. Which expression r

Alex_Xolod [135]

Answer:

Gross pay per days is 5h+8 dollars

Step-by-step explanation:

we are given

total gross pay =35h+56 dollars

total number of days =7

we know that

gross pay per day = ( total gross pay)/(total number of days)

we can plug values

gross pay per day is

$=\frac{35h+56}{7}$

we can simplify it

$=\frac{7(5h+8)}{7}$

$=5h+8$

7 0

3 years ago

Read 2 more answers

Kate has some dimes and quarters. if she has 1919 coins worth a total of $2.95, how many of each type of coin does she have?

Natalija [7]

She has 12 dimes and 7 quarters

6 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!

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

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

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.

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

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ 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