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musickatia [10]
3 years ago
5

Help??

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
7 0
The tree is 52.8 feet, since the hypotenuse is only part of the tree, you have to add 12, since then it would equal the total height of the tree.

a^2+b^2=c^2
12^2+ 39^2=c^2
144+1521= 1665
√1665
= 40.8

40.8 + 12
= 52.8 feet
Elden [556K]3 years ago
5 0

Answer:

height of the tree was 52.8 feet

Step-by-step explanation:

Height of the tree will be determined by adding broken part of the tree, lying on the ground (as the hypotenuse of the triangle) and the remaining part of the stem.

Now we apply Pythagoras theorem in the right angle triangle with the sides as 12 feet and 39 feet

⇒ Hypotenuse² = Base² + Height²

⇒ Hypotenuse² = 12² + 39²

⇒ Hypotenuse² = 144 + 1521 = 1665

⇒ Hypotenuse = √1665

                         = 40.80 feet

Now height of the tree = 12 + 40.8

= 52.8 feet

Therefore, height of the tree was 52.8 feet

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I guess it is simplifying the expression:

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The times of the runners in a marathon are normally distributed, with a mean of 3 hours and 50 minutes and a standard deviation
serious [3.7K]

The probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is 0.16 or 16%.

<h3>What is normally distributed data?</h3>

Normally distributed data is the distribution of probability which is symmetric about the mean.

The mean of the data is the average value of the given data.

The standard deviation of the data is the half of the difference of the highest value and mean of the data set.

The times of the runners in a marathon are normally distributed, with

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Refere the probabiliity table attached below. The probability of Z being inside the 1 Standard daviation of mean is 0.84.

The probability of runner selected with time less than or equal to 3 hours and 20 minutes,

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Thus, the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is 0.16 or 16%.

Learn more about the normally distributed data here;

brainly.com/question/6587992

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