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il63 [147K]
3 years ago
6

A professional golfer gets a hole-in-one 0.003% of the time they try on a hole. If the golfer plays 32,000 holes over the course

of their professional career, about how many hole-in-ones should they expect over their career?
Mathematics
2 answers:
Sauron [17]3 years ago
7 0

96 holes I think is the correct answer

olya-2409 [2.1K]3 years ago
6 0
It is 96 holes because.003 * 32000 is the number of holes that can be expected by the player.
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Craig Browning bakes cookies for the elementary school cookie sale. His chocolate chip cookies sell for $1.00 a dozen, and his o
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Answer and Step-by-step explanation:

Let

Number of chocolate chip cookies = x

Number of oatmeal brownie cookies = y

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8 0
4 years ago
The costs of different sizes of orange juice are showing in the table
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7 0
3 years ago
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PLS HELP ME ASAP FOR ALL 6!! (MUST SHOW WORK!!!) +LOTS LOTS OF POINTS!!!
Sedaia [141]
1. -4a-6=-12
add 6 to both sides
-4a-6+6=-12+6
-4a=-6
Divide by -4
a= \frac{6}{4} = 1 \frac{1}{2}

2. \frac{3}{5} z- \frac{1}{4} = \frac{17}{20}
add \frac{1}{4} to both sides but you will need to make a common denominator on the right
\frac{3}{5} z- \frac{1}{4} + \frac{1}{4} =  \frac{17}{20} +  \frac{5}{20}
\frac{3}{5}z=  \frac{11}{10}
Divide by \frac{3}{5} which means multiply by the <span>reciprocal</span>.
\frac{3}{5}* \frac{5}{3} z= \frac{11}{10} * \frac{5}{3}
z= \frac{55}{30} =1 \frac{5}{6}

3. 2 \frac{1}{2}(e-1 \frac{3}{4})=5 \frac{1}{2}
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e= \frac{9}{20}

4. 3w+7=20
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5. -.2p+.4=-1.2
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p=8

6. \frac{y}{4} - \frac{3}{5} = \frac{18}{20}
Add, made sure you change to common denominator first
\frac{y}{4} = 1 \frac{1}{4}
Multiply by 4
y=5

Hope that helps

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3 years ago
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