MATHEMATICS HIGH SCHOOL

At a high school, students can choose between three art electives, four history electives, and five computer electives. Each student can choose two electives. Which expression represents the probability that a student chooses an art elective and a history elective?

Answers

Answer 1

Answer:

Answer:

$(^3C_1* ^4C_1)/(^(12)C_2)$

Step-by-step explanation:

Given,

Art electives = 3,

History electives = 4,

Computer electives = 5,

Total number of electives = 3 + 4 + 5 = 12,

Since, if a student chooses an art elective and a history elective,

So, the total combination of choosing an art elective and a history elective = $^3C_1* ^4C_1$

Also, the total combination of choosing any 2 subjects out of 12 subjects = $^(12)C_2$

Hence, the probability that a student chooses an art elective and a history elective = $\frac{\text{Total combination of choosing an art elective and a history}}{\text{ Total combination of choosing any 2 subjects}}$

$=(^3C_1* ^4C_1)/(^(12)C_2)$

Which is the required expression.

Answer 2

Answer:

Answer: Hello!

we have:

3 art electives

4 history electives

5 computer electives

which adds to a total of 12.

If the selection is random, each elective has the same probability.

The probability of selecting an art electives is the quotient between the number of art electives and the total number of electives:

3/12

suppose that this event is true, now we need to see the probability of choosing also a history elective;

We do the same process as before, we have 4 history electives and, because we already selected 1 in the previous step, we have a total of 11 electives:

the probability now is 4/11.

Now we want to calculate the joint probability of bot events is equal to the product of their probabilities; this is:

p= (3/12)*(4/12) = (4*3)/(11*12) = 12/(11*12) = 1/11

But there is also the case where the selection is in the other order (first history and second art) so the probability is equal to

2*1/11 = 2/11

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Answers

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Answers

Answer:

1. 3 2/3

2. 4 9/10

3. 1 27/56

4. 58 2/15

5. 11 1/24

Step-by-step explanation:

1. To find the answer, add 3 1/6 and 5 9/10. First change them into improper fractions. 5 9/10 is 59/10 and 3 1/6 is 19/6. Now, you have to make the denominator (bottom number) the same number for both fractions. To do this you must find a LCM for both the number (10, 6) a LCM for these numbers is 30. Now to change the denominator to this you must times the numerator and denominator by the number needed to get the LCM for the denominator.

So 6 x 5 = 30 and 3 x 10 = 30.

19/6 x 5 = 95/30 and 59/10 x 3 = 177/30

Now add the fractions:

This equals 272/30

Simplify:

Now take 12 5/6 and turn it into a improper fraction: 77/6

Now again, you have to make the denominator to 30

77/6 x 5 = 385/30

Now subtract 385/30 - 272/30

This equals 113/30

Now turn this back into a mixed fraction:

3 23/30

The answer is 3 23/30

This process is basically the same for all the other questions.

Dr. Black is standing 13 feet from a streetlamp. The lamp is making his shadow 9 feet long. He estimates that the angle of elevation from the tip of his shadow to the top of the streetlamp is 50. To the nearest foot, the streetlamp is about _____.

Answers

Distance from tip of shadow to lamp post = 13 +9 feet = 22 ft Height of lamp post = x tan 50 = x /22 = 1.192 x = 22 * 1.192 =.26.2 ft. rounded = 26ft

Answer:

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Step-by-step explanation:

The distance between the streetlamp and the tip of his shadow is 9 + 13 = 22 ft.

A rectangle triangle is formed in which one side is 22 ft long and the other side is the streetlamp height. The angle between the 22 ft side and the hypotenuse is 50°. From tangent definition:

tan(50°) = (streetlamp height)/(22 ft)

tan(50°) * (22 ft) = streetlamp height

26 ft = streetlamp height

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Answers

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