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butalik [34]
3 years ago
13

How do I do this? Please only the best help.

Mathematics
1 answer:
MissTica3 years ago
6 0

Answer:

77.5 degrees and 102.5 degrees

Step-by-step explanation:

Given that : one angle is25 degrees greater than another. Let the two angles be represented by a and b. Since one angle is 25 degreess greater than another, then:

b = a +  25 degrees

But;

a + b = 180 degrees  (sum of angles on a straight line)

a + a + 25 degrees  = 180 degrees

2a = 180 degrees - 25 degrees

a =  77.5 degrees

b = 77.5 degrees  + 25 degrees  = 102.5 degrees

Therefore the measures of all angles formed by the lines is either 77.5 degrees  or 102.5 degrees.

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Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose f
castortr0y [4]

Answer:

Step-by-step explanation:

From the given information:

the mean (\mu) = 115 \times 20

= 2300

Standard deviation = 20 \times \sqrt{115}

Standard deviation (SD) = 214.4761

TO find:

a) P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})

P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})

P(x > 3500)= P(Z > \dfrac{1200}{214.4761})

P(x > 3500)= P(Z >5.595)

From the Z-table, since 5.595 is > 3.999

P(x > 3500)=1-0.9999

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b)

Here, the replacement time for the mean (\mu) = \dfrac{0+0.5}{2}

= 0.25

Replacement time for the Standard deviation \sigma = \dfrac{0.5-0}{\sqrt{12}}

\sigma = 0.1443

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = \sqrt{(115\times 20^2) +(114\times (0.1443)^2)}

= \sqrt{(115\times 400) +(114\times 0.02082249}

= \sqrt{(46000) +2.37376386}

= \sqrt{(46000) +(2.37376386)}

= \sqrt{46002.374}

= 214.482

Now; the required probability:

P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})

P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})

P(x > 4125) = P(Z >8.376)

P(x > 4125) =1-  P(Z

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

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3 years ago
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Answer:

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<em>If you think about it, there can only be 2 possible ways of solutions (intersection points) of an ellipse and a line.</em>

<em>1. The line will not intersect the ellipse at all, so no solution</em>

<em>2. The line will intersect the ellipse at 2 points maximum</em>

<em />

So, we can clearly see from the reasoning that the maximum number of possible solutions would be 2. The graph attached confirms this as well.

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