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PIT_PIT [208]
3 years ago
9

You received a $100 gift certificate to a clothing store. The store sells T-shirts for $15 and dress shirts for $22. You want to

spend no more than the amount of the gift certificate. You want to spend at least $90. You need at least one dress shirt. What are all the possible combinations of T-shirts and dress shirts you could buy?​
Mathematics
1 answer:
Andreas93 [3]3 years ago
7 0

Answer:

0 T, 4 D

1 T, 3 D

2T, 3D

3T, 2D

4T, 1D

5T, 1D

Step-by-step explanation:

So let's start with what we know!

Gift certificate = $100

T-shirts = $15 each

Dress shirts = $22 each

You need to buy at least one dress shirt in the process and at least spend $90

You can simply find the combinations by going in desmos and writing in:

15x + 22y <= 100

y > 1

15x + 22y >= 90

Find the combinations by looking at the integer values of x, and using only the integer values of y ( do not round them, just cut them down to the integer). Look where all the regions are shaded

The possible combinations are as follows

Note: I am going to initial t-shirts to T and dress shirts to D

0 T, 4 D

1 T, 3 D

2T, 3D

3T, 2D

4T, 1D

5T, 1D

Your welcome for the answer! If you have any questions, let me know in the comments section! If you could mark this answer as the brainliest, I would greatly appreciate it!

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{\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

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Here

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

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[It means there can be two or more tails]

Here

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

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

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

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Here

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<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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Step-by-step explanation:

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