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skelet666 [1.2K]
3 years ago
6

The length of the shadow of a tree is 24 feet, the tan P = .7.

Mathematics
1 answer:
Molodets [167]3 years ago
7 0
To find the height of the tree you need to use the basic trigonometric function known as tangent. Tangent is the equivalent of the opposite side divided by the adjacent side. If you are given the size of the angle (P), tan P = height of the tree / length of the shadow (24ft). 
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what's the length of one leg of a right triangle if the length of the hypotenuse is 25 units and the length of the other leg is
SVEN [57.7K]
20 units is the length of the other leg.
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3 years ago
You plan to invest $350 in a growth fund that has a rate of 1.5% compounded quarterly. How much money will this investment be wo
nydimaria [60]

Answer:

Hey man I know the answer to this question it would be $449.12

Step-by-step explanation: 350*1.005^50= 449.12



7 0
4 years ago
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Solve for x and then find the measure of
harkovskaia [24]

Answer:

x = 11

∠B = 80°

Step-by-step explanation:

∠A =∠B because they are corresponding angles

8x - 8 = 5x + 25

3x = 33

divide both sides of the equation by 3:

x = 11

∠B = 5(11) + 25 = 80°

7 0
3 years ago
A veterinarian’s assistant made a table that shows the animals seen in the office in one week. What is the probability that the
horsena [70]

Answer:

The probability that the pet seen was sick is 53.57%.

Step-by-step explanation:

The probability of an event <em>E</em> is the ration of the number of favorable outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

Here,

n (E) = number of favorable outcomes

N = total number of outcomes

Denote the events as follows:

<em>C</em> = the pet is a cat

<em>D</em> = the pet is a dog

<em>O </em>=<em> </em>the pet is some other animal

<em>S</em> = the pet is sick.

The data provided is summarized as follows:

n (C) = 28

n (C ∩ S) = 3

n (D) = 42

n (B ∩ S) = 4

n (O) = 24

n (O ∩ S) = 8

Compute the probability that a cat was sick as follows:

P(C\cap S)=\frac{n(C\cap S)}{n(C)}=\frac{3}{28}

Compute the probability that a dog was sick as follows:

P(D\cap S)=\frac{n(D\cap S)}{n(D)}=\frac{4}{42}=\frac{2}{21}

Compute the probability that another animal was sick as follows:

P(O\cap S)=\frac{n(O\cap S)}{n(O)}=\frac{8}{24}=\frac{1}{3}

Compute the probability that the pet seen was sick as follows:

P (S) = P (C ∩ S) + P (D ∩ S) + P (O ∩ S)

       =\frac{3}{28}+\frac{2}{21}+\frac{1}{3}\\=\frac{9+8+28}{84}\\=\frac{45}{84}\\=0.5357

Thus, the probability that the pet seen was sick is 53.57%.

8 0
3 years ago
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Paul can type fast but 20 words less than Jennifer a minute.
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4 years ago
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