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3 years ago

7

Jon performs 3 transformations on a pentagon. First, he reflects it over the x-axis. Then, he translates it 3 units to the left

and 4 units up. Finally, he reflects the image over the line x = 1. Are the original and transformed images congruent? Justify your answer.

1 answer:

Morgarella [4.7K]3 years ago

8 0

yes; Reflections and translations are rigid motions. Rigid motions preserve the size of the original image.

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Mrs. Tart takes a survey in class. She discovers 60% of her students prefer using pens over pencils in class.

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20 students

Step-by-step explanation:

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You are trying to find the distance from Saint Augustine to Jacksonville. What would be the most appropriate unit of measure to

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. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

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Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total <em>4+5+6 = 15 bulbs</em>. If we want to select 3 randomly there are K ways of doing this, where K is the<em> combination of 15 elements taken 3 at a time </em>

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since <em>the events are disjoint</em>, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, <em>supposing there is no replacement</em>, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$

Since there are no replacement, the probability of taking a second non 75-W bulb is now $\frac{8}{14}=0.5714$

Following this procedure 5 times, we find the probabilities

$\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}$

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

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3 years ago

Every time I use a piece of scrap paper, I crumple it up and try to shoot it inside the recycling bin across the room. I'm prett

valentinak56 [21]

Using the binomial distribution, it is found that there is a 0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

With 5 shoots, the probability of making at least one is $\frac{211}{243}$ , hence the probability of making none, P(X = 0), is $\frac{232}{243}$ , hence:

$(1 - p)^5 = \frac{232}{243}$

$\sqrt[5]{(1 - p)^5} = \sqrt[5]{\frac{232}{243}}$

1 - p = 0.9908

p = 0.0092

Then, with 6 shoots, the parameters are:

n = 6, p = 0.0092.

The probability that at least two of them make it inside the recycling bin is:

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

In which:

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

Then:

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

$P(X = 0) = C_{6,0}.(0.0092)^{0}.(0.9908)^{6} = 0.9461$

$P(X = 1) = C_{6,1}.(0.0092)^{1}.(0.9908)^{5} = 0.0527$

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.9461 + 0.0527 = 0.9988

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9988 = 0.0012$

0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

More can be learned about the binomial distribution at brainly.com/question/24863377

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2 years ago

you have to answer 3 essay questions for an exam. there are 6 essays to check choose from. how many different groups of 3 essays

Elis [28]

According to the fundemintal counting princable the answer is 18.

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3 years ago