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geniusboy [140]
3 years ago
10

Identify four over five as proper, improper or mixed

Mathematics
2 answers:
Wittaler [7]3 years ago
8 0
Proper because improper would be 6 over 5 and mixes would be 1 and 1 over 6
Marizza181 [45]3 years ago
7 0
It would be proper because he numerator is not a higher value than five
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Please Help, and show work..
azamat
Hi there! So the the new price is $93, and went up from $55. Let's subtract the numbers. 93 - 55 is 38. The change is 38. To find the percent markup, we can write and solve a proportion. Set it up like this:

38/55 = x/100

This is because the amount of change goes above the original (55), and x represents the percent out of 100. We are looking for the markup. Let's cross multiply the values. 100 * 38 is 3,800. 55 * x is 55x. 3,800 = 55x. Now, divide each side by 55 to isolate the x. 55x/55 cancels out. 3,800/55 is 69.0909 or 69 when rounded to the nearest whole number. There. The percent of markup is approx. 69%.
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Answer:

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Step-by-step explanation:

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Math help, BOTH of these please!
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#1 : ( 3 , - 2 )

~~

#2 : ( - 1 , 4 )

~~

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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}

8 0
2 years ago
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