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borishaifa [10]
3 years ago
14

The weight V of an object on Venus varies directly with its weight E on Earth. A person weighing 120 lb on Earth would weigh 106

lb on Venus. How much would a person weighing 150 lb on Earth weigh on Venus?
Mathematics
2 answers:
sukhopar [10]3 years ago
5 0
We have
120lbs Earth................................106lbs Venus
150lbs Earth...................................?Venus

Venus weight=150*106/120=132.50 lbs
svlad2 [7]3 years ago
4 0

Answer:

The person weighing 150 lb on Earth will weight 132.5 lb on Venus.

Step-by-step explanation:

The weight of object on Venus varies directly with its weight on the Earth.

w_v\propto w_e..(1)

\frac{w_v}{w_e}=k=constant

w_v= Weight on Venus

w_e= Weight on Earth

If the weight of person on Earth =w_e=120 lb

If the weight of that person on Venus=w_v=106 lb

And,the weight of another person on Earth =w_e'=150 lb

The weight of another person on Venus=w_v'=?

\frac{w_v}{w_e}=\frac{w_v'}{w_e'} (From 1)

w_v'=\frac{w_v\times w_e'}{w_e}=\frac{106 lb\times 150 lb}{120 lb}

w_v'=132.5 lb

The person weighing 150 lb on Earth will weight 132.5 lb on Venus.

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If a coin is tossed three times, find probability of getting
Assoli18 [71]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

‣ A coin is tossed three times.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

‣ The probability of getting,

1) Exactly 3 tails

2) At most 2 heads

3) At least 2 tails

4) Exactly 2 heads

5) Exactly 3 heads

{\large{\textsf{\textbf{\underline{\underline{Using \: Formula :}}}}}}

\star \: \tt  P(E)= {\underline{\boxed{\sf{\red{  \dfrac{ Favourable \:  outcomes }{Total \:  outcomes}  }}}}}

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

★ When three coins are tossed,

then the sample space = {HHH, HHT, THH, TTH, HTH, HTT, THT, TTT}

[here H denotes head and T denotes tail]

⇒Total number of outcomes \tt [ \: n(s) \: ] = 8

<u>1) Exactly 3 tails </u>

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

\therefore  \sf Probability_{(exactly  \: 3 \:  tails)}  =  \red{ \dfrac{1}{8}}

<u>2) At most 2 heads</u>

[It means there can be two or one or no heads]

Here

• Favourable outcomes = {HHT, THH, HTH, TTH, HTT, THT, TTT} = 7

• Total outcomes = 8

\therefore  \sf Probability_{(at \: most  \: 2 \:  heads)}  =  \green{ \dfrac{7}{8}}

<u>3) At least 2 tails </u>

[It means there can be two or more tails]

Here

• Favourable outcomes = {TTH, TTT, HTT, THT} = 4

• Total outcomes = 8

\longrightarrow   \sf Probability_{(at \: least \: 2 \:  tails)}  =  \dfrac{4}{8}

\therefore  \sf Probability_{(at \: least \: 2 \:  tails)}  =   \orange{\dfrac{1}{2}}

<u>4) Exactly 2 heads </u>

Here

• Favourable outcomes = {HTH, THH, HHT } = 3

• Total outcomes = 8

\therefore  \sf Probability_{(exactly \: 2 \:  heads)}  =  \pink{ \dfrac{3}{8}}

<u>5) Exactly 3 heads</u>

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

\therefore  \sf Probability_{(exactly \: 3 \:  heads)}  =  \purple{ \dfrac{1}{8}}

\rule{280pt}{2pt}

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Answer:

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