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navik [9.2K]
3 years ago
13

A rocking horse has a weight limit of 60pounds. What percentage of the weight limit is 33 pounds? What percentage of the weight

limit is 114 pounds? What weight is 95% of the limit?
Mathematics
1 answer:
bixtya [17]3 years ago
4 0

Answer:

The percentage is <u>55%</u> when weight limit is 33 pounds.

The percentage is <u>190%</u> when weight limit is 114 pounds.

The 95% of the limit weighs <u>57</u> pounds.

Step-by-step explanation:

Given:

Weight limit of rocking horse = 60 pounds.

We need to find:

a) What percentage of the weight limit is 33 pounds.

b)What percentage of the weight limit is 114 pounds.

c) What weight is 95% of the limit.

Now Solving for a we get;

Weight limit = 33 pounds

Now percentage of the weight limit can be calculated by dividing given weight limit with total weight limit and then multiplying by 100 we get;

framing in equation form we get;

percentage of the weight limit = \frac{33}{60}\times100 = 55\%

Hence The percentage is <u>55%</u> when weight limit is 33 pounds.

Now Solving for b we get;

Weight limit = 114 pounds

Now percentage of the weight limit can be calculated by dividing given weight limit with total weight limit and then multiplying by 100 we get;

framing in equation form we get;

percentage of the weight limit = \frac{114}{60}\times100 = 190\%

Hence The percentage is <u>190%</u> when weight limit is 114 pounds.

Now Solving for c we get;

Percentage Weight = 95%

Now Amount of weight limit can be calculated by multiplying Percentage weight limit with total weight limit and then Dividing by 100 we get;

framing in equation form we get;

percentage of the weight limit = \frac{95}{100}\times60 = 57 \pounds

Hence The 95% of the limit weighs <u>57</u> pounds.

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sertanlavr [38]

Answer:

a. 72

b. 816

c. 570

d. 30

Step-by-step explanation:

Given the graph is a bell - shaped curve. So, we understand that this is a normal distribution and that the bell - shaped curve is a symmetric curve.

Please refer the figure for a better understanding.

a. In a normal distribution, Mean = Median = Mode

Therefore, Median = Mean = 72

b. We have to know that 68% of the values are within the first standard deviation of the mean.

i.e., 68% values are between Mean  Standard Deviation (SD).

Scores between 63 and 81 :

Note that 72 - 9 = 63 and

72 + 9 = 81

This implies scores between 63 and 81 constitute 68% of the values, 34% each, since the curve is symmetric.

Now, Scores between 63 and 81 =  

= 68 X 12 = 816.

That means 816 students have scored between 63 and 81.

c. We have to know that 95% of the values lie between second Standard Deviation of the mean.

i.e., 95% values are between Mean  2(SD).

Note that 90 = 72 + 2(9) = 72 + 18

Also, 54 = 63 - 18.

Scores between 54 and 90 totally constitute 95% of the values. So, Scores between 72 and 90 should amount to  of the values.

Therefore, Scores between 72 and 90 =  

= 570.

That is a total of 570 students scored between 72 and 90.

d. We have to know that 5 % of the values lie on the thirst standard Deviation of the mean.

In this case, 5 % of the values lie between below 54 and above 90.

Since, we are asked to find scores below 54. It should be 2.5% of the values.

So, Scores below 54 =  

= 2.5 X 12 = 30.

That is, 30 students have scored below 54.

Step-by-step explanation:

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