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harkovskaia [24]
3 years ago
8

Which equation represents the value of x? x=100−y2−−−−−−−√ x=10+y2 x=10−y x=y2+100−−−−−−−√ Right triangle A B C with angle B as

the right angle. A C is equal to 10. B C is equal to y. A B is equal to x.

Mathematics
1 answer:
Nostrana [21]3 years ago
8 0

Answer:

Therefore the required value of x,

x=\sqrt{(100-y^{2})}

Step-by-step explanation:

Given:

ΔBC is a Right Angle Triangle at ∠ B = 90°

As ∠ B = 90° , AC will be the Hypotenuse

AC = 10 = Hypotenuse

BC = y = Longer leg ( say )

AB = x = Shorter leg ( say )

To Find :

x = ?

Solution:

In Right Angle Triangle Δ ABC , By Pythagoras Theorem we get

(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}

Substituting the given values we get

10^{2}= x^{2}+y^{2} \\\\x^{2}=100-y^{2} \\\\\textrm{square rooting on both the side we get}\\\\x=\sqrt{(100-y^{2})}\\\therefore x=\sqrt{(100-y^{2})}\ \textrm{ which is the required value of x}

Therefore the required value of x,

x=\sqrt{(100-y^{2})}

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10 months ago
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Suppose that Kevin can choose to get home from work by car or bus. When he chooses to get home by car, he arrives home after 7 p
astraxan [27]

Answer:

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm = 0.838

Step-by-step explanation:

Let the probability that Kevin arrives home after 7 pm be P(L)

Probability that Kevin uses the bus = P(B)

Probability that Kevin uses the car = P(C)

Probability of arriving home after 7 pm if the car was taken = P(L|C) = 4% = 0.04

Probability of arriving home after 7 pm if the bus was taken = P(L|B) = 15% = 0.15

The bus is cheaper, So, he uses the bus 58% of the time.

P(B) = 58% = 0.58

P(C) = P(B') = 1 - P(B) = 1 - 0.58 = 0.42

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm = P(B|L)

The conditional probability P(A|B) is given mathematically as

P(A|B) = P(A n B) ÷ P(B)

Hence, the required probability, P(B|L) is given as

P(B|L) = P(B n L) ÷ P(L)

But we do not have any of P(B n L) and P(L)

Although, we can obtain these probabilities from the already given probabilities

P(L|C) = 0.04

P(L|B) = 0.15

P(B) = 0.58

P(C) = 0.42

P(L|C) = P(L n C) ÷ P(C)

P(L n C) = P(L|C) × P(C) = 0.04 × 0.42 = 0.0168

P(L|B) = P(L n B) ÷ P(B)

P(L n B) = P(L|B) × P(B) = 0.15 × 0.58 = 0.087

P(L) = P(L n C) + P(L n B) = 0.0168 + 0.087 = 0.1038 (Since the bus and the car are the two only options)

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm

= P(B|L) = P(B n L) ÷ P(L)

P(B n L) = P(L n B) = 0.087

P(L) = 0.1038

P(B|L) = (0.087/0.1038) = 0.838150289 = 0.838

Hope this Helps!!!

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