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3 years ago

What can you conclude about dilations from your observations in question 5

Mathematics

1 answer:

Over [174]3 years ago

5 0

Answer:

The angles remain the same for all values of n. After a dilation, the angle measurements for the image and the preimage are always equal.

Step-by-step explanation:

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Before ordering numbers, what from would you need to make them all into

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To order the numbers, start with the lowest (first) and work your way up (last). This is referred to as "Ascending Order." Consider climbing a mountain.

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3 years ago

-20 2 -6 -18 4 -4 put them from least to greatest

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-20 -18 -6 -4 2 4 this is the answer

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Hilda was simplifying some numerical expressions and made each of the following sequences of calculations. Name the mathematical

Galina-37 [17]

Answer:

Problem 1:

5 · (−4/3) · (2/5)

= (−4/3) · 5 · (2/5) Conmutative Property of Multiplication

= (−4/3) · (5 · (2/5)) Associative Property of Multiplication

= (−4/3) · (2/1) = −8/3 = −2*2/3 Multiplied fractions and extracted common factor

Problem 2:

17 + 29 + 3+ 1

= 17 + 3 + 29 + 1 Conmutative Property of Addition

= (17 + 3) + (29 + 1) Associative Property of Addition

= 20 + 30 Added groups

= 50 Added terms

Step-by-step explanation:

For the Problem 1:

In the first step, Hilda applied the Conmutative Property of Multiplication, because she changed the order of the numbers in the product

In the second step, she applied the Associative Property of Multiplication, because she agrouped the product of 5 and 2/5 to perform it sepparately

In the third step, she calculated the product of the fractions -4/3 and 2/1, then she extracted 2 as a common factor to express the fraction as -2*2/3

For the Problem 2:

In the first step, Hilda applied the Conmutative Property of Addition, because she changed the order of the numbers in the sum

In the second step, she applied the Associative Property of Addition, because she associated the addition of 17 and 3 and the addition of 29 and 1, to calculate them in groups.

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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.

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3 years ago

2, 2, 3, 4, 5, 6 what is the probability of spinning a 2? (in percentage) PLEAZ HELP

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Answer:

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Step-by-step explanation:

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3 years ago

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