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dezoksy [38]
3 years ago
9

There are 4 colas, 1 ginger ale, 7 root beers, and 6 cherry sodas in a cooler. What are the odds of against choosing a ginger al

e?
Mathematics
2 answers:
Schach [20]3 years ago
8 0
Ok, the last one I gave you the wrong answer. So, I'll try again. I apologize if I'm wrong

Add the all of the pops up except for the ginger ale.

4+7+6+6=23

There is 1 ginger ale, so the probability is 1 to 23 or 1/23


Paraphin [41]3 years ago
7 0
One out of eighteen drinks.
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Opposite lateral faces of a rectangular pyramid are congruent.

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Two neighboring stores sell apples. Sunny Market sells apples in groups of 5 for $3.20. Happy Mart sells apples in groups of 2 f
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Plz help me with this math and also explain
jasenka [17]

Step-by-step explanation:

<h2>[1]</h2>

  • SI = $250
  • Rate (R) = 12\sf \dfrac{1}{2} %
  • Time (t) = 4 years

\longrightarrow \tt { SI = \dfrac{PRT}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 12\cfrac{1}{2} \times 4}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times \cfrac{25}{2} \times 4}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 25 \times 2}{100} } \\

\longrightarrow \tt { 250 = \dfrac{P \times 50}{100} } \\

\longrightarrow \tt { 250 \times 100 = P \times 50} \\

\longrightarrow \tt { 25000 = P \times 50} \\

\longrightarrow \tt { \dfrac{25000}{50} = P } \\

\longrightarrow \underline{\boxed{ \green{ \tt { \$ \; 500 = P }}}} \\

Therefore principal is $500.

<h2>__________________</h2>

<h2>[2]</h2>

  • 2/7 of the balls are red.
  • 3/5 of the balls are blue.
  • Rest are yellow.
  • Number of yellow balls = 36

Let the total number of balls be x.

→ Red balls + Blue balls + Yellow balls = Total number of balls

\longrightarrow \tt{ \dfrac{2}{7}x + \dfrac{3}{5}x + 36 = x} \\

\longrightarrow \tt{ \dfrac{10x + 21x + 1260}{35} = x} \\

\longrightarrow \tt{ \dfrac{31x + 1260}{35} = x} \\

\longrightarrow \tt{ 31x + 1260= 35x} \\

\longrightarrow \tt{ 1260= 35x-31x} \\

\longrightarrow \tt{ 1260= 4x} \\

\longrightarrow \tt{ \dfrac{1260 }{4}= x} \\

\longrightarrow \underline{\boxed{  \tt { 315 = x }}} \\

Total number of balls is 315.

A/Q,

3/5 of the balls are blue.

\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}x} \\

\longrightarrow \tt{ Balls_{(Blue)} =\dfrac{3 }{5}(315)} \\

\longrightarrow \tt{ Balls_{(Blue)} = 3(63)} \\

\longrightarrow \underline{\boxed{ \green {\tt { Balls_{(Blue)} = 189 }}}} \\

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