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3 years ago

The sum of two consecutive odd integers is 28. What are the integers?

1 answer:

4 0

The correct option is B): "Let x = 1st integer. Let x + 2 = 2nd integer."
Given that "The sum of two consecutive odd integers is 28."

Now the difference between two odd numbers is 2. So this means that one odd number has to be greater than the other by 2.

Let the smaller number be 'X' therefore the bigger number will be 'x+2' thus x + (x+2) = 28

therefore, x + x + 2 = 28 now we subtract "2" from both the sides
therefore, 2x = 26 now dividing throughout by 2 we get:

x = 13 therefore, x + 2 = 13+2
= 15

therefore, the numbers are 13 AND 15

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Answere 4 and 6 PLEASE show work

max2010maxim [7]

I think 4 is 3 pages plus 1/4=3 1/4

and 6 is 5+1/3= 5 1/3

hope this helpa

6 0

2 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = 0.0083.

3 0

3 years ago

Read 2 more answers

What is value of the digit 5 in the number 75

ycow [4]

First, think of your places. You have the ones places, tens places, hundreds places, and so on.

The first number starting from the right is the ones, and as you keep going left, the value of each given digit becomes higher.

Since 5 is in the ones place, its value would be just 5. If it were in the tens place, it would be 50. If it were in the hundreds place, it would be 500, and so on.

Think of it this way;

Ones is just one. If a number is in the 'ones' place, its value would be a single digit. If it were in the tens place, its value would be two digits.
That's how it would be for each place going left.
Every number you move to the left, its value gains a one.

So here's an example:
5555

The value of 5 in the ones place "5555" is simply 5.

In the tens place, you end up adding one zero, so the value of the second five to the left would be, "50"

So with that said, the value of the digit 5 in the number 75 is 5.

Haha, hope this cleared up any confusion, and have a wonderful day! :)

7 0

3 years ago

Read 2 more answers

Pls help xDDDDD thx!

prohojiy [21]

Answer:

-25

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

13. f(x) = x + 5 a. f(4) b. f(7) c. f(-3) d. f(0) e. f(2.4) f. f(2/3)