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Dahasolnce [82]
3 years ago
11

A box is filled with 9 blue cards, 4 green cards, and 3 yellow cards. A card is chosen at random from the box. What is the proba

bility that it is a blue or a yellow card? Write answer as a fraction.
Mathematics
2 answers:
VLD [36.1K]3 years ago
6 0
12/16 which would reduce to 6/8 which can also reduce further to 3/4! :) so the final answer is 3/4.
sp2606 [1]3 years ago
4 0
Start by finding the total sum of cards in the box.

9+4+3= 16

Now what we are looking for is how many cards we are looking for to pull (yellow or blue).

9+3 =12

We could pull 12 cards that would be yellow or blue out of a box of 16 cards.

Our answer would be 12/16 simplified to 3/4.
You might be interested in
How many sets of three consecutive integers are there in which the sum of the three integers equals their product?
skad [1K]

Answer:

3

Step-by-step explanation:

since the 3 integers are consecutive, we are dealing with x, x+1, x+2.

and their sum is the same as their product :

x + (x + 1) + (x + 2) = x(x + 1)(x + 2)

3x + 3 = x(x² + 3x + 2) = x³ + 3x² + 2x

x³ + 3x² - x - 3 = 0

this is a polynomial of third degree.

and as such it has 3 solutions.

of course, it could be that some of them are the same or are even in the realm of complex numbers (i = sqrt(-1)), but usually these 3 solutions are different real numbers.

I tried x=1 just to see, and, hey, it is a solution for this equation.

x = 1 means that the other 2 consecutive integers are 2 and 3.

and indeed, 1+2+3 = 1×2×3 = 6.

now it is easier to find the other 2 solutions, as a zero solution can be expressed as a factor of the whole expression.

for x = 1 the factor term is (x - 1), as this term is then turning 0, when x = 1.

I can divide the main expression by this factor and then analyze the quotient about the other 2 solutions.

x³ + 3x² - x - 3 : x - 1 = x² + 4x + 3

- x³ - x²

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

0 4x² - x

- 4x² - 4x

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

0 + 3x - 3

- 3x - 3

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

0 0

so, the original expression can be written as

(x² + 4x + 3)(x - 1).

now we need to find the 2 zero solutions for x²+4x+3

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 4

c = 3

so,

x = (-4 ± sqrt(4² - 4×1×3))/(2×1) =

= (-4 ± sqrt(16 - 12))/2 = (-4 ± sqrt(4))/2 =

= (-4 ± 2)/2 = -2 ± 1

x1 = -2 + 1 = -1

x2 = -2 - 1 = -3

so, we have the additional solutions :

-1 0 1

-3 -2 -1

-1 + 0 + 1 = -1×0×1 = 0

-3 + -2 + -1 = -3×-2×-1 = -6

and there we have it fully proven :

there are 3 different sets of 3 consecutive integers with the same sum as product.

4 0
2 years ago
17. Atlanta, Georgia, receives an average of 27 inches of precipitation per year, and there has been 9 inches so far this year.
kirill115 [55]

Answer:

  9 + p ≤ 27; p ≤ 18

Step-by-step explanation:

"At or below" means "less than or equal to." Only one answer choice incorporatest that symbol into the inequality expression — the one shown above.

5 0
3 years ago
Read 2 more answers
Prove for any positive integer n, n^3 +11n is a multiple of 6
suter [353]

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

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

Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

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

Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

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

If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

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

In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

6 0
3 years ago
Read 2 more answers
Help please and ty ! (no its not 388)
jeka94
It is 484 I knwonir
5 0
3 years ago
Read 2 more answers
Because of stormy weather a pilot flying at 35,000 ft descends 8,000 ft.
babymother [125]

Answer:

35,000 - 8000=27,000 altitude

6 0
1 year ago
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