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3 years ago

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

Mathematics

1 answer:

iogann1982 [59]3 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

Now,

Total no. of trial when room shall light

=10-3

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

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Vitek1552 [10]

Answer:

Its 100 percent B

Step-by-step explanation:

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3 years ago

Find the gcf of <br> 24, 14, 21

kobusy [5.1K]

Answer:

Step-by-step explanation:

All the numbers provided can have a factor of 1.

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1*24=24

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3 years ago

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Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

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3 years ago

Hector is dividing students into groups for an HR wants to divide the boys and girls so that each group has the same number of t

Roman55 [17]

Answer:

7 groups, 3 boys and 8 girls

Step-by-step explanation:

Let total number of groups be $x$

and total hikers in one group be $y$

if number of girls and boys in every group is same, then

$\frac{21}{x} + \frac{56}{x} = y$

$\frac{21 + 56}{x} = y\\x*y = 77$ ....(1)

from (1) $x$ will either be 11 or 7

for $x = 11$ the values won't be real.

For $x = 7\\y = \frac{77}{7} \\y = 11$

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mezya [45]

Is it multiple choice? If so, leave the answer choices.

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2 years ago

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