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3 years ago

Find the x is qs bisects pqr and pqr=82

Mathematics

3 answers:

slavikrds [6]3 years ago

5 0

if ∡PQR = 82°, and the ray QS bisects it, it cuts ∡PQR into two equal halves, ∡PQS and ∡RQS, each of which is then 82/2, or 41°.

$\bf \stackrel{\measuredangle PQS}{10x+1}=41\implies 10x=40\implies x=\cfrac{40}{10}\implies x=4$

Cerrena [4.2K]3 years ago

3 0

Answer:

$x=4$

Step-by-step explanation:

We have been given that ray QS bisects angle PQR and measure of angle PQR is 82 degrees. We are asked to find the value of x for our given diagram.

Since ray QS bisects angle PQR, so measure of each angle formed by ray QS will be half the measure of angle PQR.

$m\angle PQS=\frac{m\angle PQR}{2}$

$10x+1=\frac{82}{2}$

$10x+1=41$

$10x+1-1=41-1$

$10x=40$

$\frac{10x}{10}=\frac{40}{10}$

$x=4$

Therefore, the value of x is 4.

7289Keke3 years ago

0 0

lnbbgdjdidi

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The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine

salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

$f(x)=\frac{e^{\frac{-x}{1000} }}{1000}$

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

b. P(1000<x<2000) = 0.2325

c. P(x<1000) = 0.6321

d. 105.4 hours

Step-by-step explanation: Probability Density Function is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = $\int\limits^b_a {P(x)} \, dx$

Then, for the electronic component, probability will be:

P(a<x<b) = $\int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx$

P(a<x<b) = $\frac{1000}{1000}.e^{\frac{-x}{1000} }$

P(a<x<b) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

a. For a component to last more than 3000 hours:

P(3000<x<∞) = $e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}$

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = $e^{-3}$

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = $e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}$

P(1000<x<2000) = $e^{-2}-e^{-1}$

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = $e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}$

P(0<x<1000) = $e^{-1}-1$

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

0.1 = $1-e^\frac{-x}{1000}$

$-e^{\frac{-x}{1000} }=-0.9$

${\frac{-x}{1000} }=ln0.9$

-x = -1000.ln(0.9)

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10% of the components will have failed at 105.4 hours.

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Tpy6a [65]

Answer:

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O y = 0.5(x2 -2x - 16) becomes y = 0.5(-16) = -8 (reject)

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