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3 years ago

Ray BA and BC are opposite rays, and Ray BE bisects

Mathematics

1 answer:

Monica [59]3 years ago

5 0

Ray BE bisects ∠ABD.

According to picture, this is the solution. In case the question is complete.

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. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$

Since there are no replacement, the probability of taking a second non 75-W bulb is now $\frac{8}{14}=0.5714$

Following this procedure 5 times, we find the probabilities

$\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}$

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

3 0

3 years ago

What are the solutions of the system?

svetlana [45]

Answer:

Option C. No solution is the right answer.

Step-by-step explanation:

Here the given equations are y = x²+2x+3 -----(1)

and y = 4x-2 -------(2)

Now we substitute the value of y from equation 2 into 1.

x²+2x+3 = 4x-2

x²+2x+3-2x = 4x-2-2x

x²+3 = 2x-2

x²+3-2x = 2x-2x-2

x²-2x+3 = -2

x²-2x+5 = 0

Then value of $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

$=\frac{+2\pm \sqrt{4-4\times 1\times 5}}{2}$

$=\frac{2\pm \sqrt{4-20}}{2}$

Since in this solution √(-20) is not defined. Therefore there is no solution.

3 0

3 years ago

Read 2 more answers

Amy has to write a paper for school that has more than 2,878 words. She has already written 1,037 words and writes at a rate

densk [106]

Answer:

Step-by-step explanation:

263x = words per day

add on 1,037 because she already wrote them

263x + 1,037 > 2,878

7 0

3 years ago

help The screen in a theatre is 22 ft high and is positioned 10 ft above the floor, which is flat. The first row of seats is 7 f

Stella [2.4K]

(a.)
Let's say α is the angle that subtends from the top of the screen to horizontal eye-level.
Let β be the angle that subtends from the bottom of the screen to horizontal eye-level.

tanα = (22 + 10 - 4) / x = 28/x
α = arctan(28/x)

tanβ = (10 - 4) / x = 6/x
β = arctan(6/x)

Ɵ = α - β
Ɵ = arctan(28/x) - arctan(6/x)

(b.)
tanƟ = tan(α - β) = (tanα - tanβ) / (1 + tanα tanβ)
tanƟ = (28/x - 6/x) / [1 + (28/x)(6/x)]
tanƟ = (22/x) / [1 + (168/x²)]
tanƟ = 22x / (x² + 168)
Ɵ = arctan[22x / (x² + 168)]

4 0

3 years ago

8.97 in word form and expanded form

avanturin [10]

8+.90+.07
This the answer because you have to make it as an addition way.
Hope I helped.

4 0

3 years ago

Read 2 more answers