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Deffense [45]
3 years ago
7

Triangle ABC is similar to triangle PQR, as shown below:

Mathematics
2 answers:
dalvyx [7]3 years ago
8 0
Let's see here... well, since the first letter in the ratio is from the first triangle, then, the first one in the equal ratio is going to have to be that as well, and the ratio of c:a wouldn't work, since c was already used. By process of elimination, it's gotta be b:q.
Montano1993 [528]3 years ago
8 0

we know that

If triangle ABC is similar to triangle PQR

then

the corresponding sides are proportional

so

\frac{AC}{PR}= \frac{AB}{PQ}= \frac{BC}{QR}

substitute the values

\frac{b}{q}= \frac{c}{r}= \frac{a}{p}

\frac{c}{r}=\frac{b}{q}

\frac{c}{r}=\frac{a}{p}

therefore

<u>the answer is the option</u>

b:q


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If you roll two six-faced dice together, you will get 36 possible outcomes. 4 pts 1. List all possible outcomes of the experimen
bezimeni [28]

Given:

Two dice are rolled together.

Total number of possible outcomes.

To find:

The list of total possible outcomes.

The probability of getting a sum of 11 in these outcomes.

The probability of getting a sum less than or equal to 4.

The probability of getting a sum of 13 or more.

Solution:

If two dice are rolled together, then the total number of possible outcomes is 36 and list of total possible outcomes is

S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Sum of 11 in these outcomes = {(5,6),(6,5),(6,6)} = 3

The probability of getting a sum of 11 in these outcomes is

P(\text{sum=11})=\dfrac{3}{36}

P(\text{sum=11})=\dfrac{1}{12}

Therefore, the probability of getting a sum of 11 in these outcomes is \dfrac{1}{12}.

Sum less than or equal to 4 = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} = 6

The probability of getting a sum less than or equal to 4 is

P({sum\leq 4})=\dfrac{6}{36}

P({sum\leq 4})=\dfrac{1}{6}

Therefore, the probability of getting a sum less than or equal to 4 is \dfrac{1}{6}.

Sum of 13 or more = empty set because maximum sum is 12.

The probability of getting a sum of 13 or more is

P(sum\geq 13)=\dfrac{0}{36}

P({sum\geq 13})=0

Therefore, the probability of getting a sum of 13 or more is 0.

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Artyom0805 [142]

Answer:

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