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Fed [463]
3 years ago
8

Find the value of x and explain how you found it. Make sure to cite the problem and share what you see before you explain your s

olution. In your explanation, include how you know that these triangles are similar. You may use a postulate such as AA, SSS, SAS, Triangle Proportionality Theorem, or a creative alternative. Justify why you were able to use the postulate/theorem and give the equation you used to find x.

Mathematics
1 answer:
klemol [59]3 years ago
6 0

Answer:

THE VALUE OF X = 3 .

Step-by-step explanation:

From the Δ RQP and Δ TSP ;

⇒ ∠  TPS = ∠ RPQ       ( since both are common angle )

⇒ RQ ≈ TS                    ( parallel lines )

⇒ ∠ PRQ = ∠ PST            ( Corresponding angle )

∴ Δ RQP ≅  Δ TSP          ( BY ASA RULE )

Now ,

 \frac{RP}{TP}  = \frac{QP}{PS}

\frac{9 + X}{9}  = \frac{6 + 2}{6}

⇒ X = 3

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-0.183 as a fraction in the simplest form??
IRISSAK [1]
-0.183 = -0.183 / 1

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3 years ago
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Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let <em>X</em> = number of vehicles that pass the inspection.

The probability of the random variable <em>X</em> is <em>P (X) = 0.70</em>.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

                                    =(0.70)^{3}\\=0.343

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

                                   =1-0.343\\=0.657

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

                         = P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

                              + P (Only 3rd vehicle passes)

                       =(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

                                                       + P (0 vehicles passes)

                                              =0.189+(0.30\times0.30\times0.30)\\=0.216

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let <em>X</em> = all 3 vehicle passes and <em>Y</em> = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

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Answer:

Step-by-step explanation:

2,57 divided by 1.5 equals

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