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WITCHER [35]
4 years ago
11

Triangles PQR and XYZ are similar.

Mathematics
1 answer:
laiz [17]4 years ago
7 0

Answer:

Option A. 5.25 units

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

so

\frac{PR}{XZ}=\frac{QR}{YZ}

substitute the given values

\frac{7}{XZ}=\frac{4}{3}

solve for XZ

XZ=7(3)/4\\XZ=5.25\ units

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The first one is B and the second is D
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Solve for x. –2(6 – x) = 3x A. –3 B. 12 C. 3 D. –12
Katen [24]

-2(6-x)= 3x, the solution is  find the value of x,  you must apply distributive property, tho the first term,  -12+2x= 3x⇒ 3x-2x= -12⇒ x = -12, the  answer correct is option D

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Consider the continuous random variable x, which has a uniform distribution over the interval from 110 to 150. The probability t
Virty [35]

Answer:

The probability that x will take on a value between 120 and 125 is 0.14145

Step-by-step explanation:

For uniform distribution between a & b

Mean, xbar = (a + b)/2

Standard deviation, σ = √((b-a)²/12)

For 110 and 150,

Mean, xbar = (150 + 110)/2 = 130

Standard deviation, σ = √((150-110)²/12 = 11.55

To find the probability that x will take on a value between 120 and 125

We need to standardize 120 & 125

z = (x - xbar)/σ = (120 - 130)/11.55 = - 0.87

z = (x - xbar)/σ = (125 - 130)/11.55 = - 0.43

P(120 < x < 125) = P(-0.87 < x < -0.43)

We'll use data from the normal probability table for these probabilities

P(120 < x < 125) = P(-0.87 < x < -0.43) = P(z ≤ -0.43) - P(z ≤ -0.86) = 0.33360 - 0.19215 = 0.14145

Hope this Helps!!!

3 0
3 years ago
A game is played with a spinner on a circle, like the minute hand on a clock. The circle is marked evenly from 0 to 100, so, for
zheka24 [161]

Answer:

The probability is 1/2

Step-by-step explanation:

The time a person is given corresponds to a uniform distribution with values between 0 and 100. The mean of this distribution is 0+100/2 = 50 and the variance is (100-0)²/12 = 833.3.

When we take 100 players we are taking 100 independent samples from this same random variable. The mean sample, lets call it X, has equal mean but the variance is equal to the variance divided by the length of the sample, hence it is 833.3/100 = 8.333.

As a consecuence of the Central Limit Theorem, the mean sample (taken from independant identically distributed random variables) has distribution Normal with parameters μ = 50, σ= 8.333. We take the standarization of X, calling it W, whose distribution is Normal Standard, in other words

W = \frac{X - \mu}{\sigma} = \frac{X - 50}{8.333} \simeq N(0,1)

The values of the cummulative distribution of the Standard Normal distribution, lets denote it \phi , are tabulated and they can be found in the attached file, We want to know when X is above 50, we can solve that by using the standarization

P(X > 50) = P(\frac{X-50}{8.33} > \frac{50-50}{8.33}) = P(W > 0) = \phi(0) = 1/2

Download pdf
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4 years ago
Eva sees a dolphin 3.2 meters below sea level and a bird 47/10 meters above sea level. Which of the following expressions repres
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7.9 meters apart from each other
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