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Tanya [424]
3 years ago
15

a line has a slope of -2 and includes the points (-1,r) and (2,4). write the value of r in the empty box

Mathematics
1 answer:
guajiro [1.7K]3 years ago
4 0

Answer:

r= 10

Step-by-step explanation:

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How do you solve 36 times <img src="https://tex.z-dn.net/?f=%5Csqrt%7B3%7D" id="TexFormula1" title="\sqrt{3}" alt="\sqrt{3}" ali
pantera1 [17]

Answer:

  62.3538

Step-by-step explanation:

There is nothing to solve. If you need a decimal value, you can use a calculator or table of square roots.

8 0
3 years ago
A spinner has equal regions numbered 1 through 20. What is the probability that the spinner will stop on an odd number or a mult
klio [65]

Answer:

<em>Probability that the spinner will stop on an odd number or a multiple of 5 is </em><em>0.6</em>

Step-by-step explanation:

Probability = \frac{Required outcomes}{Total possible outcomes}

We are given the equal regions numbered from 1 through 20 which means that our total possible outcomes are 20

<em>Total possible outcomes: 20</em>


<em>Outcomes that spinner will stop on an odd number, n(Odd): 10</em>

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Probability of spinner stoping on Odd number:

P(Odd) = \frac{n(Odd)}{Total} = \frac{10}{20} = \frac{1}{2} = 0.5


Outcomes that spinner will stop on a multiple of 5, n(5): 4

5, 10, 15, 20

Probability of spinner stoping on multiple of 5:

P(5) = \frac{n(5)}{Total} = \frac{4}{20} = \frac{1}{5} = 0.2

Odd numbers which are a multiple of 5 are: 5 and 15

So,

P(Odd and 5) = \frac{2}{20}=\frac{1}{10}=0.1

Thus Probability of spinner stopping at odd number or a multiple of 5 becomes:

P(Odd or 5) = P(Odd) + P(5) - P(Odd and 5) = 0.5 + 0.2 - 0.1 = 0.6

7 0
3 years ago
Fraction with a denominator of 10 can be written as an equivalent fraction with
AVprozaik [17]

Answer:

9/10 is 90%

Step-by-step explanation:

You just make it a percentage

5 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
I NEED THIS ASAP, PLEASE HELP!!!
WINSTONCH [101]

Answer: 22.5 ; 5 ; 14

Step-by-step explanation:

Given the dataset:

{20,22,23,24,26,26,28,29,30}

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 9

Q1 = 1/4 (9 + 1)th term

Q1 = 1/4(10) = 2.5

We average the 2nd and 3rd term:

(22 + 23) / 2

45 / 2 = 22.5

B) The interquartile range(IQR) of the dataset :

{62,63,64,65,67,68,68,68,69,74}

IQR = Q3 - Q1

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 10

Q1 = 1/4 (10 + 1)th term

Q1 = 1/4(11) = 2.75 term

We take the average of the 2nd and 3rd term:

(63 + 64) / 2

45 / 2 = 63.5

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 10

Q3 = 3/4 (10 + 1)th term

Q3 = 3/4(11) = 8.25 term

We take the average of the 8th and 9th term:

(68 + 69) / 2

137 / 2 = 68.5

IQR = Q3 - Q1

IQR = 68.5 - 63.5

IQR = 5

C) give the dataset :

{7,8,8,9,10,12,13,15,16}

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 9

Q3 = 3/4 (9 + 1)th term

Q3 = 3/4(10) = 7.5 term

We take the average of the 7th and 8th term:

(13 + 15) / 2

28 / 2 = 14

7 0
3 years ago
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