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kap26 [50]
3 years ago
10

A spinner is divided into many different color sectors. Some sectors are larger than others. The spinner was spun 300 times. The

results where tallied and written in the table below.
Red:93 Blue:59 Green:105 Yellow:43


Part A: What is the probability that the next spin will land on green?


Part B: What is the probability that the next spin will land on either green or red?
Mathematics
1 answer:
Furkat [3]3 years ago
4 0
Solutions 

We know that <span>a spinner was divided into many different color sectors. Some sectors are larger than others. The spinner was spun 300 times. The results were tallied up and were written in a table. 

</span><span>Red:93

Blue:59

Green:105

Yellow:43
</span>
The spinner landed the highest on green. To find <span>the probability that the next spin will land on green we have add all the numbers and simplify them.  

93 + 59 + 105 + 43 = 300 

Out of 300 the spinner landed to green 105 times. We have the fraction 105/300. 

105/300 can be reduced. To reduce the fraction we need to the GCF of 105 and 300. The GCF is 15. 

</span>We can reduce the fraction by dividing the numerator and denominator by the GCF = 15. 

105 ÷ 15 = 7 

300 ÷ 15 = 20 

Our new fraction is 7/20 

The <span>probability that the next spin will land on green is 7/20. 
</span>≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
<span>
To get the answer to part (B) we have to add the results of red and green then simplify. 

93 + 105 = 198 

The fraction is 198/300. </span><span>198/300 can be reduced. To reduce the fraction we need to the GCF of 105 and 300. The GCF is 6. 

</span>We can reduce the fraction by dividing <span>the numerator and denominator by the GCF = 6.

</span>198 ÷ 6 = 33

300 ÷ 6 = 50<span>

The new fraction is 33/50 

</span>The probability that the next spin will land on either green or red is <span>33/50. 
</span>
Hope this helps 
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Answer:

x = 3, y = -4

Step-by-step explanation by substitution:

Solve the following system:

{4 x + 3 y = 0 | (equation 1)

5 y + 53 = 11 x | (equation 2)

Express the system in standard form:

{4 x + 3 y = 0 | (equation 1)

-(11 x) + 5 y = -53 | (equation 2)

Swap equation 1 with equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

4 x + 3 y = 0 | (equation 2)

Add 4/11 × (equation 1) to equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+(53 y)/11 = -212/11 | (equation 2)

Multiply equation 2 by 11/53:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+y = -4 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{-(11 x)+0 y = -33 | (equation 1)

0 x+y = -4 | (equation 2)

Divide equation 1 by -11:

{x+0 y = 3 | (equation 1)

0 x+y = -4 | (equation 2)

Collect results:

Answer: {x = 3 , y = -4

_____________________________________

Solve the following system:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for x:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Isolate terms with x to the left hand side.

Subtract 3 y from both sides:

{4 x = -3 y

5 y + 53 = 11 x

Hint: | Solve for x.

Divide both sides by 4:

{x = -(3 y)/4

5 y + 53 = 11 x

Hint: | Perform a substitution.

Substitute x = -(3 y)/4 into the second equation:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for y:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Isolate y to the left hand side.

Subtract 53 - (33 y)/4 from both sides:

{x = -(3 y)/4

(53 y)/4 = -53

Hint: | Solve for y.

Multiply both sides by 4/53:

{x = -(3 y)/4

y = -4

Hint: | Perform a back substitution.

Substitute y = -4 into the first equation:

Answer: {x = 3 , y = -4

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