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3 years ago

Solve the equation for x

Mathematics

1 answer:

KengaRu [80]3 years ago

7 0

You can do this 2 ways, one is knowing the special right triangles called triplets. An example of a triplet would be 3-4-5 where the largest value is the hypotenuse. In this example, the other 2 sides are 3 and 4 and the missing side is the hypotenuse, so using the knowledge of triplets, we can find that the hypotenuse is 5.

algebraically, we can use the Pythagorean theorem; a^2+b^2=c^2 where a and b are the opposite and adjacent and where c is the hypotenuse. We can then substitute in the values to get : 3^2 + 4^2 = c^2. When simplified, we get 9+16=c^2. Simplify further, and get 25=c^2. Now we Square root both sides to solve for C. Finally, we get C=5. Your missing side is 5.

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3 years ago

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 33% of all homeowners are ins

Lina20 [59]

Answer:

(a) The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b) The most likely value for X is 1.32.

Step-by-step explanation:

The random variable X is defined as the number among the four homeowners who have earthquake insurance.

The probability that a homeowner has earthquake insurance is, p = 0.33.

The random sample of homeowners selected is, n = 4.

The event of a homeowner having an earthquake insurance is independent of the other three homeowners.

(a)

All the statements above clearly indicate that the random variable X follows a Binomial distribution with parameters n = 4 and p = 0.33.

The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b)

The most likely value of a random variable is the expected value.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected value of X as follows:

$E(X)=np$

$=4\times 0.33\\=1.32$

Thus, the most likely value for X is 1.32.

(c)

Compute the probability that at least two of the four selected have earthquake insurance as follows:

P (X ≥ 2) = 1 - P (X < 2)

= 1 - P (X = 0) - P (X = 1)

$=1-{4\choose 0}\ (0.33)^{0}\ (1-0.33)^{4-0}-{4\choose 1}\ (0.33)^{1}\ (1-0.33)^{4-1}\\\\=1-0.20151121-0.39700716\\\\=0.40148163\\\\\approx 0.4015$

Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.

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3 years ago

TriangleABC has three sides with these lengths: AB=9, BC=40, and CA=41, What is the value of cos C?

Otrada [13]

Answer:

9² = 40² + 41² - 2(40)(41) cos(C)

81 = 1600 + 1681 - 3280 cos(C)

81 = 3281 - 3280 cos(C)

-3200 = -3280 cos(C)

cos(C) = 3200 / 3280

cos(C) = 40 / 41

Step-by-step explanation:

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3 years ago