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3 years ago

1. Find the distance from point P to QS

Mathematics

1 answer:

12345 [234]3 years ago

4 0

Answer:

5.7 units

Step-by-step explanation:

The distance from point P to QS is the distance from point P (1, 1) to the point of interception R(-3, 5).

Use distance formula to calculate distance between P and R:

$PR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$P (1, 1) = (x_1, y_1)$

$R(-3, 5) = (x_2, y_2)$

Plug in the values into the formula.

$PR = \sqrt{(-3 - 1)^2 + (5 - 1)^2}$

$PR = \sqrt{(-4)^2 + (4)^2}$

$PR = \sqrt{16 + 16}$

$PR = \sqrt{32}$

$PR = 5.7 units$ (to nearest tenth)

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Select from the drop-down menus to correctly complete each statement.

Studentka2010 [4]

Answer:

To find the absolute value of a number Find the distance from number to zero on a number line . The absolute value of a number is always Positive or zero .

Step-by-step explanation:

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3 years ago

Question 61 pts Richard has been given a 12-question multiple-choice quiz in his history class. Each question has three answers,

Brut [27]

Answer:

0.946 = 94.6% probability that he will answer at least 2 questions correctly.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of other questions. 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.

12 questions.

This means that $n = 12$

Each question has three answers, of which only one is correct.

This means that $p = \frac{1}{3} = 0.3333$

Assuming that Richard guesses on all 12 questions, find the probability that he will answer at least 2 questions correctly.

Either he answers less than 2 questions correctly, or he answers at least 2 questions correctly. The sum of the probabilities of these outcomes is decimal 1. So

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

We want $P(X \geq 2)$ . So

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = x) = C_{12,0}.(0.3333)^{0}.(0.6667)^{12} = 0.008$

$P(X = x) = C_{12,1}.(0.3333)^{1}.(0.6667)^{11} = 0.046$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.008 + 0.046 = 0.054$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.054 = 0.946$

0.946 = 94.6% probability that he will answer at least 2 questions correctly.

3 0

3 years ago

A large study of German children of Caucasian descent found that 46% have brown hair and that 25% have freckles. Nonetheless, yo

STatiana [176]

Answer:

Having brown hair and having freckles are not necessarily independent events

Step-by-step explanation:

We are given the following information in the question:

Percentage of children having brown hairs = 46%

Percentage of people children having freckles = 25%

But it cannot be concluded that

German children of Caucasian descent are brown‑haired with freckles =

$0.46\times 0.25 = 0.115 = 11.5\%$

This is not always true because of the following reasons:

Brown hair and having freckles are not necessarily independent events.

We can not assume that event of a child having brown hair is independent of the event of child having freckles. The events could be dependent as well.

The above will only be true when event of a child having brown hair is independent of the event of child having freckles.

If the events are independent then,

$P(A \cap B) = P(A)\times P(B)$