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miss Akunina [59]
3 years ago
8

Probability help please? In a certain instant lottery game, the chances of a win are stated as "1 in 19." Express the indicated

degree of likelihood as a probability value between 0 and 1 inclusive. (round to three decimal places) I am confused and need a step by step instructions. Please help
Mathematics
1 answer:
Vanyuwa [196]3 years ago
5 0
The question is basically asking to find the chances in percentage. It is first given in fraction form, which is 1/19. This is approximately 0.053, once divided.

1/19 (One divided by 19)=0.0526

0.0526 rounded is about 0.053 (to three decimal places).

Answer: About 0.053
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Mr. Santiago can buy light fixtures in packages of 12 and light bulbs in packages of 9. He bought the fewest number of light fix
liq [111]

Answer:

  • 3 and 4 packages

Step-by-step explanation:

We need to find the LCM of 12 and 9.

  • 9 = 3*3
  • 12 = 2*2*3

<u>Their LCM is:</u>

  • LCM(12, 9) = 2*2*3*3 = 36

Number of light fixtures and light bulbs is 36.

<u>Number of packages will be:</u>

  • 36/12 = 3 and
  • 36/9 = 4 respectively.
6 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
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Lady_Fox [76]
57,682
I got this by putting the numbers in their place 

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What is x\3 so solve for x if it equals 90
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X= 270
simplify the equation using cross multiplication
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Write an expression to show how many meters are equivalent to x centimeters
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1 meter is 100 centimeters so...

x meters are equal to 100x centimeters
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