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Vesnalui [34]
3 years ago
11

Explain how you can use a model to find 6x17

Mathematics
1 answer:
VashaNatasha [74]3 years ago
6 0
Well, I'd say make a 6x17 array:
▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄
▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄
▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄
▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄
▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄
▄▄▄▄▄▄▄▄▄▄▄▄▄▄▄ 

By the way, 6x17=102
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What’s the answer for this question??
vladimir1956 [14]
The answer is for w is 16
7 0
3 years ago
Please solve this question faaaast​
Kazeer [188]

Answer:

300-400

Step-by-step explanation:

The first step is finding the total of the data we have. So, we take 5 + 10 + 15 + 20 + 25 + 15 + 10 which equals 80.

The median is the middle point of all the data. If it's an odd number, you can calculate the median with the equation (n+1) / 2, plugging in the total amount of data for n.

When it's an even number though, there is no one middle point since the data splits evenly in 2, so we have to use 2 equations: n/2 & (n/2) + 1. Then, we find the average of those two data points. (Although, you don't have to do that for this particular question).

When we plug 80 in for n in both of these equations, we get 40 and 41.

To where this is in the question, we have to count up from the bottom of the chart. 1-5 is below 100, 6-15 is 100-200, 16-30 is 200-300, and 31-50 is 300-400.

Since 40 and 41 are between 31 and 50, the answer is 300-400.

Hope this helps! :)

6 0
2 years ago
Answer the question below
ankoles [38]
Answer:a
explanation:
6(1) cm2
6 cm 2


good luck!
7 0
2 years ago
<img src="https://tex.z-dn.net/?f=3%20%5Cfrac%7B3%7D%7B4%7D%20%20%5Ctimes%20%5Cfrac%7B2%7D%7B5%7D%20" id="TexFormula1" title="3
Gwar [14]

Answer:

1 1/2

Step-by-step explanation:

8 0
3 years ago
A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple
tester [92]

Answer:  The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation:  We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.

Therefore, the probability of event A is given by

P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

4 0
3 years ago
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