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Olenka [21]
3 years ago
5

According to the Centers for Disease Control, the probability that a randomly selected citizen of the United States has hearing

problems is 0.151. The probability that a randomly selected citizen of the United States has vision problems is 0.093. Can we compute the probability of randomly selecting a citizen of the United States who has a hearing problem or vision problem by adding these probabilities? Why or why not?
Mathematics
1 answer:
laila [671]3 years ago
5 0
I think the answer is yes you can because they are both different things.

What I mean by that is that the United Stated as .151% of its people with hearing problems, and .093% with vision problems. So really adding them together just mixes it saying that .244% of people in the united states have either a vision or a hearing problem.

It doesn’t make it any more or any less people by adding them together. Therefore it’s not any more or any less probable to pick one or the other.

Hopes this helps.
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The hypotenuse of a right angled triangle is 2√13 cm . If the smaller side is increased by 2 cm and the larger side is increased
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<em><u>Statement:</u></em>

The hypotenuse of a right angled triangle is 2√13 cm. If the smaller side is increased by 2 cm and the larger side is increased by 3 cm, the new hypotenuse will be √117 cm.

<em><u>To find out:</u></em>

The length of the larger side of the right angled triangle.

<em><u>Solution:</u></em>

Let us consider x as the smaller side and y as the larger side.

Then, in the right angled triangle,

x² + y² = (2√13)² ...(I) [By Pythagoras Theorem]

Now, if the smaller side is increased by 2 cm, then the smaller side will be (x + 2).

And if the larger side is increased by 3 cm, then the larger side will be (y + 3).

Then, in the new right angled triangle,

(x + 2)² + (y + 3)² = (√117)² [By Pythagoras Theorem]

or, x² + 2 × 2 × x + 2² + y² + 2 × 3 × y + 3² = (√117)²

or, x² + 4x + 4 + y² + 6y + 9 = (√117)²

or, x² + y² + 4x + 6y + 13 = (√117)²

Now, put the value of x² + y² from equation (I),

or, (2√13)² + 4x + 6y + 13 = (√117)²

or, (2 × 2 × √13 × √13) + 4x + 6y + 13 = (√117 × √117)

or, 52 + 4x + 6y + 13 = 117

or, 4x + 6y = 117 - 52 - 13

or, 4x + 6y = 52

or, 4x = 52 - 6y

or, x = \frac {(52 - 6y)}{4} ...(II)

Now, put the value of x of equation (II) in (I),

x² + y² = (2√13)²

or, \frac {(2704-624y +36y²)}{16} + y² = 52

or, \frac {(2704-624y +36y² + 16y²)}{16}= 52

or, 52y²-624y + 2704 = 52 × 16

or, 52y² - 624y + 2704 - 832 = 0

or, 52y² - 624y + 1872 = 0

or, 52(y² - 12y + 36) = 0

or, y²-12y +36 = 0 ÷ 52

or, y²-12y +36 = 0

or, (y)² - 2 × 6 × y + (6)² = 0

or, (y - 6)² = 0

or, y - 6=0

or, y = 6

We have taken y as the length of the larger side of the right angled triangle.

So, the length of the larger side is 6 cm.

<em><u>Answer:</u></em>

6 cm

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If you have any query, feel free to ask.

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