Who is Albert Einstein and i need for a essay so far i have

Pythagorean Theorem and its Converse <br><br> I need number 6

Black_prince [1.1K]

Answer:

x = 14.02 units

Step-by-step explanation:

Finding unknown value using Pythagorean theorem:

This is an isosceles trapezium. The two perpendiculars divide the trapezium into two congruent triangles and a rectangle. Let the base of the triangle by 'y'. We can find the value of 'y' using Pythagorean theorem.

In ΔABC,

AC² = AB² + BC²

19² = 17 ² + y ²

361 = 289 + y ²

361 - 289 = y ²

y ² = 72

y =√72

y = 8.49

x = 31 - 8.49 - 8.49

x = 14.02

6 0

2 years ago

About 19% of the population of a large country is hopelessly romantic. If two people are randomly selected, what is the probab

Aleksandr-060686 [28]

Answer:

3.61% probability that both are hopelessly romantic.

34.39% probability at least one is hopelessly romantic

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are hopelessly romantic, or they are not. The probability of a person being hopelessly romantic is independent from other people. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

19% of the population of a large country is hopelessly romantic.

This means that $p = 0.19$

Two people are randomly selected

This means that $n = 2$

What is the probability both are hopelessly romantic?

This is P(X = 2). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,2}.(0.19)^{2}.(0.81)^{0} = 0.0361$

3.61% probability that both are hopelessly romantic.

What is the probability at least one is hopelessly romantic?

$P(X \geq 1) = P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{2,1}.(0.19)^{1}.(0.81)^{1} = 0.3078$

$P(X = 2) = C_{2,2}.(0.19)^{2}.(0.81)^{0} = 0.0361$

$P(X \geq 1) = P(X = 1) + P(X = 2) = 0.3078 + 0.0361 = 0.3439$

34.39% probability at least one is hopelessly romantic

7 0

3 years ago

6a = 175 can you help me I really need this grade

Mila [183]

Answer:

divide 175 by 6 then write a = 29.1666666

6 0

3 years ago

Read 2 more answers

When the domain of a function has an infinite number of values, the range always has an infinite number of values. True or false

iragen [17]

Answer:

Thus, the statement is False!

Step-by-step explanation:

When the domain of a function has an infinite number of values, the range may not always have an infinite number of values.

For example:

Considering a function

$f(x) = 5$

Its domain is the set of all real numbers because it has an infinite number of possible domain values.

But, its range is a single number which is 5. Because the range of a constant function is a constant number.

Therefore, the statement ''When the domain of a function has an infinite number of values, the range always has an infinite number of values'' is FALSE.

Thus, the statement is False!

3 0

3 years ago

A standard number cube numbered one through six on each side is rolled five times. What is the probability of landing on a three

Sav [38]

Answer:

1/243 ≈ 0.004 (which is a low probability)

Explanation:

The event landing on a three or four on a rolling is independent of the event of landing on a three or four on any other rolling.

Thus, for independent events you calculate the joint probability as the product of the probabilities of each event:

P(A and B and C and D and E) = P(A) × P(B) × P(C) × P(D) × P(E)

Here, P(A) = P(B) = P(C) = P(D) = P(E) = probability of landing on a three or four.

Now to calculate the probability of landing on a three or four, you follow this reasoning:

Positive outcomes = three or four

Number of positive outcomes = 2

Possible outcomes = one, two, three, four, five, or six

Number of possible outcomes = 6

Probability = numer of positive outcomes / numer of possible outcomes

Probability = 2/6 = 1/3

Finally, probability of landing on a three or four on all five rolls: (1/3)⁵ = 1/243 ≈ 0.004 ← answer

5 0

4 years ago