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3 years ago

A person is 6 feet tall and his shadow is 10 feet if a trees shadow is 25 feet long how big is the tree

Mathematics

1 answer:

Sidana [21]3 years ago

6 0

Answer:

The tree is 15 feet tall

Step-by-step explanation:

6 feet tall - 10 feet shadow

x feet tall - 25 feet shadow

x=(25*6)/10=15 feet

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How do I work this out?

tia_tia [17]

Answer:

Option 2, 3(5y-11) - 2

Hope this helps!

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3 years ago

Match the following items.

Pani-rosa [81]

This problem can be solved using probability, the equation of the probability of an event A is P(A)= favorable outcomes/possible outcomes. The interception of two probable events is P(A∩B)= P(A)P(B).

There are 12 black marbels, 10 red marbles, and 18 white marbels, all the same size. If two marbles are drawn from the jar without being replaced.

The total of the marbles is 40.

If two marbles are drawn from the jar without being replaced, what would the probability be:

1. of drawing two black marbles?

The probability of drawing one black marble is (12/40). Then, the probability of drawing another black marble after that is (11/39) due we drawing one marble before.

P(Black∩Black) = (12/40)(11/39) = 132/1560, simplifying the fraction:

P(Black∩Black) = 11/130

2. of drawing a white, then a black marble?

The probability of drawing one white marble is (18/40). Then, the probability of drawing then a black marble after that is (12/39) due we drawed one marble before.

P(White∩Black) = (18/40)(12/39) = 216/1560, simplifying the fraction:

P(White∩Black) = 9/65

3. of drawing two white marbles?

The probability of drawing one white marble is (18/40). Then, the probability of drawing another white marble after that is (17/39) due we drawed one marble before.

P(White∩White) = (18/40)(17/39) = 306/1560, simplifying the fraction:

P(White∩White) = 51/260

4. of drawing a black marble, then a red marble?

The probability of drawing one black marble is (12/40). Then, the probability of drawing then a red marble after that is (10/39) due we drawed one marble before.

P(Black∩Red) = (12/40)(10/39) = 120/1560, simplifying the fraction:

P(Black∩Red) = 1/13

4 0

3 years ago

Solve the system of linear equations. separate the x- and y- values with a coma. 20x=-58-2y

lianna [129]

The best way to solve is by using elimination method.
20x = -58 - 2y
17x = -49 - 2y
Multiply second equation by -1
20x = -58 - 2y
-17x = 49 + 2y
Add equations.
3x = -9
Divide.
x = -3
Plug in -3 into one of the equations.
17(-3) = -49 - 2y
-51 = -49 - 2y
Add 49 to both sides.
-2 = -2y
Divide.
1 = y
So your solution is (-3, 1).
I hope this helps love! :)

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3 years ago

(-3,4) is one of many solutions to the inequality: <br><br> 2x+y ≥ -2<br><br> True or false

Blababa [14]

Answer:

True

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$