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3 years ago

Which sequences of transformations applied to shape I prove that shape I is similar to shape II?

Mathematics

1 answer:

Paraphin [41]3 years ago

7 0

Answer:

A. The reflection across x axis ,fallowed by a reflection across the y axis ,and then a dilation by a scale factor of 0.5

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What is the equation of the blue line

VladimirAG [237]

Answer:

what line

Step-by-step explanation:

7 0

3 years ago

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A basketball team has 10 players. Before each game, the coach picks 2 of those players to carry the team's

labwork [276]

The different group of 2 players that the coach can pick is 45 groups.

<h3>Selection of groups</h3>

The selection of groups of 2 player can be done using the method of combination.

<h3>Different groups of 2 players</h3>

The different group of 2 players that the coach can pick from the 10 players is calculated as follows;

n = 10C2

$n = \frac{10!}{(10-2)! 2!} = \frac{10 \times 9\times 8!}{8! \times 2} = 45$

Thus, the different group of 2 players that the coach can pick is 45 groups.

Learn more about combinations here: brainly.com/question/25821700

6 0

2 years ago

"5 to the second power", what does the exponent 2 tell us to do with the<br> base 5?

masha68 [24]

Answer:

The exponent 2 tells us that we must multiply 2 five's together.

So we have: 5^2 = 5 x 5

Let me know if this helps!

7 0

3 years ago

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PLS HELP ASAP I WILL IVE BRAINLIST ANSWER!!!!!Leon verified that the side lengths 21, 28, 35 form a Pythagorean triple using thi

Juliette [100K]

Answer:

Leon is correct. (Option 1)

Step-by-step explanation:

Given that Leon verified that the side lengths 21, 28, 35 form a Pythagorean triple using this procedure.

Step 1: Find the greatest common factor of the given lengths: 7

Step 2: Divide the given lengths by the greatest common factor: 3, 4, 5

Step 3: Verify that the lengths found in step 2 form a Pythagorean triple.

we have to explain whether or not Leon is correct.

As, 3,4,5 forms a Pythagorean triplet i.e satisfies the Pythagoras theorem

$Hypotenuse^2=Base^2+Perpendicular^2$

⇒ $5^2=3^2+4^2$

Let a, b, c forms a Pythagorean triplet

$a^2+b^2=c^2$

Multiplied by 4 on both sides

⇒ $4a^2+4b^2=4c^2$

⇒ ${2a}^2+{2b}^2={2c}^2$

Hence, we say 4a, 4b and 4c also forms a Pythagorean triplet.

∴ multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple.

Hence, Leon is correct.

4 0

3 years ago

Read 2 more answers

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

3 years ago