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azamat
3 years ago
13

The formula of the area of annulus correct to one decimal place

Mathematics
2 answers:
S_A_V [24]3 years ago
7 0

Answer:

Step-by-step explanation:

It’s 4

Arturiano [62]3 years ago
6 0
The answer:
it is 4 my friend
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Whats 1+1 Plz help enjoy the points hehe
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Answer:

2

Step-by-step explanation:

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In a certain Algebra 2 class of 29 students, 13 of them play basketball and 7 of them play baseball. There are 4 students who pl
Inessa [10]

Answer:

16/29

Step-by-step explanation:

P(A∪B) = P(A) + P(B) - P(A∩B)

P(basketball or baseball) = P(basketball) + P(baseball) - P(both)

= (13/29) + (7/29) - (4/29)

= 16/29

The probability that a randomly chosen student plays either sport is 16/29.

3 0
3 years ago
HElPPPPP NEED QUiCk whats the answer for 2+2=____<br> a.5<br> b.7<br> c.4<br> d.3
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Answer:

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2 years ago
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14.7x + 4x = 0*(46y+39y) Y=14, Solve for X
Dovator [93]
14.7x + 4x = 0*(49y+39y) y = 14

18.7x = 0(686 + 546)

18.7x = 0

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4 0
3 years ago
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}

8 0
1 year ago
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