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Rudiy27
3 years ago
13

What is the mode? 8 8 6 8 6

Mathematics
2 answers:
____ [38]3 years ago
4 0
Your question mode is 8 because its reapting more times
Rashid [163]3 years ago
3 0

Answer:8

Step-by-step explanation:

It appears the most in the set of data

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A light bulb consumes 3120 watt-hours in 3 days and 6 hours. How many watt-hours does it consume per day?
matrenka [14]

2850*24/(4*24+18)

The light bulb consumes 600 watt-hours per day.

6 0
3 years ago
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
942 ÷ 7 = ? A. 134 R4 B. 136 C. 134 R2 D. 133 R1
zimovet [89]

Answer:


Step-by-step explanation:

134

5 0
3 years ago
Read 2 more answers
Helllllllllllllllpppppppppp
Leona [35]

Answer:

bird, sea lion, fish

Step-by-step explanation:

-0.5 sea lion, -1.2 fish, 4.25 bird

8 0
2 years ago
Read 2 more answers
Find the area of the shaded region.
muminat

Answer:

It's 38.1

Step-by-step explanation:

I'm not the best at explaining these type of things but I hope this helps you out

5 0
2 years ago
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