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Alex_Xolod [135]

3 years ago

10

your friend is Graphing the point (5,3) the first step is to start at the organ and go 5 spaces. which statement about his metho

d is true

1 answer:

Lena [83]3 years ago

6 0

Answer:

Moving 5 spaces

Step-by-step explanation:

Although your friend is correct in moving 5 spaces, they don't specify which direction they move in. They could be moving left, right, up, or down.

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What is 200×300?<br><br>A. 60,000<br><br>B. 5,000<br><br>C. 50,000<br><br>D. 6,000

GrogVix [38]

Answer:

A. 60,000

Step-by-step explanation:

The answer is A because 2 × 3 is 6 then you add the 4 0's. So it would be 60,000

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2 years ago

Which type of transformation preserves the symmetry of an even function f(x) but does not preserve the symmetry of an odd functi

Alexus [3.1K]

The type of transformation that preserves the symmetry of an even function f(x) but does not preserve the symmetry of an odd function g(x) is vertical translation.

<h3>What is vertical translation?</h3>

Vertical translation of a graph is done by moving the base graph up or down in the y-axis direction. Each point on a graph is moved k units vertically to translate the graph by that many units.

The vertical translation is the movement of the curve along the y-axis by a certain number of units without altering the function's shape or domain.

The function's form is preserved in the case of vertical translation.

Learn more about Vertical Translation here:

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7 0

2 years ago

3 + 3 x 3 - 3 + 3 =?

SCORPION-xisa [38]

12, 3x3 = 9, +3 =12-3=9+3=12

3 0

3 years ago

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Wiil mark bianleast plz plz hlpe

Andrews [41]

Answer:

18 since 12 x 2 = 24 so 9 x 2 = 18

Step-by-step explanation:

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3 years ago

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Seventy-two percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of th

Anton [14]

Answer:

a) 0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) 0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) 0.064 = 6.4% probability that 7 of them are discovered.

Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

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.

a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

$P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312$

Probability of having an emergency locator and not being discovered:

20% of 28%. So

$P(A cap B) = 0.2*0.28 = 0.056$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.056}{0.5312} = 0.105$

0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

$P(A) = 1 - 0.5312 = 0.4688$

Probability of not having an emergency locator and being discovered:

34% of 72%. So

$P(A \cap B) = 0.34*0.72 = 0.2448$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2448}{0.4688} = 0.522$

0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?

p is the probability of being discovered with the emergency recorder:

0.5312 probability of having the emergency recorder.

Probability of having the emergency recorder and being located:

66% of 72%. So

$P(A \cap B) = 0.66*0.72 = 0.4752$

Probability of being discovered, given that it has the emergency recorder:

$p = P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4752}{0.5312} = 0.8946$

This question asks for P(X = 7) when $n = 10$ . So

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

$P(X = 7) = C_{10,7}.(0.8946)^{7}.(0.1054)^{3} = 0.064$

0.064 = 6.4% probability that 7 of them are discovered.

8 0

3 years ago