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vesna_86 [32]
2 years ago
8

Five bulbs of which three are defective are to be tried in two bulb points in a dark room

Mathematics
1 answer:
iogann1982 [59]2 years ago
8 0

Answer:

D 7

Step-by-step explanation:

Total no. of bulb= 5

no. of defected bulb= 3

no. of not defected bulb=2

Total no. of bulb combination = 5C2

=5!/2!(5-2)!

= 5!/2!3!

= 5×4×3×2×1/2×1×3×2×1

=120/12

=10

( since a room can be lighted with one bulb also)

total no. of bulb combination when room shall not light = 3C2

3!/2!(3-2)!

= 3!/2! 1!

= 3×2×1/2×1×1

= 6/2

=3

Now,

Total no. of trial when room shall light

=10-3

=7

Hence, number of trial when the room shall be lighted is 7 which is option d

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At a factory, a mechanic earns $17.25 an hour. The president of the company earns 6 3/4 times as much for each hour he works. Th
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Find how much each earns

pres =6 and 3/4 times mec pay=17.25 times 6.75=116.4375, about $116.44
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add them

$17.25+$116.4275+$12.9375=146.615
round
$146.62
7 0
2 years ago
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A skier moves 100.0 m horizontally, and then travels another 35.0 m uphill at an angle of 35.0º above the horizontal. What is th
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Because there isnt 90 degrees angle in this triangle (triangle - starting point - point where he starts climbing- ending point) we will use cosine law to find magnitude of displacement. For cosine law we need 2 sides of triangle and angle between them which is exactly what is given.

a^2 = b^2 + c^2 - 2*b*c*cos(alpha)
after expressing values we get:
a^2 = 10000 + 1225 + 5734
a = 130,2 meters

to calculate angle we again use cosine law but now our unknown variable is angle alpha. our sides we will use are 100 meters and 130,2 meters because we need angle between them.

cos(alpha) = (b^2 + c^2 - a^2)/(2*b*c)
cos(alpha) = 0.98
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8 0
2 years ago
michael wittry has been investing inhis roth IRA retirement account for 20 years. Two years ago his account was worth $215,658.
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Let's say

a = 215,658

ok... so.. he first lost 1/3 of that

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\\\\\\
\textit{then he \underline{gained} }\frac{1}{2}\textit{ of }\frac{2a}{3}\qquad \qquad \cfrac{2a}{3}+\cfrac{\frac{2a}{3}}{2}\implies \cfrac{\frac{2a}{3}}{\frac{2}{1}}
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\cfrac{2a}{3}+\cfrac{2a}{3}\cdot \cfrac{1}{2}\implies \cfrac{2a}{3}+\cfrac{2a}{6}\implies \cfrac{4a+2a}{6}
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4 0
2 years ago
6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the
MaRussiya [10]

Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

a_1 the horizontal distance between the first observer and the ballonist

a_2 the horizontal distance between the second observer and the ballonist

\alpha _1 and \alpha _2 the angles of elevation meassured by each observer

S the area of the triangle formed with the observers and the ballonist

So, the area of a triangle is the length of its base times its height.

S=a*h (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

S=S_1+S_2 (equation 2)

S_1=a_1*h

But we can write S_1 in terms of \alpha _1, like this:

\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}

And for S_2 will be the same:

S_2=\frac{h^{2} }{\tan(\alpha _2)}

Replacing in the equation 2:

S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})

And replacing in the equation 1:

h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}

So, we can replace all the known data in the last equation:

h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

6 0
2 years ago
The list shows numbers in order from least to greatest. Negative 34, negative 2 and three-fourths, blank, negative 1.5, 0, 2.3,
blsea [12.9K]

Options

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  • -2
  • -1\dfrac{9}{10}
  • -1

Answer:

(B) -2

Step-by-step explanation:

Given the list

-34.-2\dfrac{3}{4} ,x, -1.5,0,2.3,10,25\dfrac15

This list has been arranged in order from the least to the greatest.

We are required to find an Integer that can be inserted on the blank line (I have used x) in the list.

An integer is defined as a positive or negative whole number.

Out of the given options, only -1 and -2 are integers. However:

  • -1 is greater than -1.5

Therefore, it cannot be our result.

The correct answer is -2.

3 0
2 years ago
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