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Sati [7]
4 years ago
13

One fourth of the candy bars were not chocolate. There were 45 chocolate candy bars. One-fifth of the chocolate candy bars conta

ined nuts. How many nonchocolate candy bars were there? How many chocolate candy bars contained nuts?
Mathematics
1 answer:
xeze [42]4 years ago
4 0
1/4 =non chocolate bar
3/4 = chocolate bar

45 ÷ 3 = 15
So 1/4 = 15 candy bars

And 1/5 candy bars contained nuts...

45 x 1/5 = 9 chocolate bars contained nuts

The answer is ...
15 non chocolate bars
And
9 chocolate candy bars contained nuts
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Naddik [55]
14 divided by 17 is 14/17.

Hope this helps~
5 0
4 years ago
33
olga2289 [7]

Answer:

$1.61

Step-by-step explanation:

Tax paid = 7% of $23

= 0.07*23

= $1.61

8 0
3 years ago
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Which expression shows the result of applying the distributive property to -2(1/3x - 1/5) ?
Alexeev081 [22]

Answer:

A. -2/3x - 2/5

Step-by-step explanation:

-2 x 1/3x = -2/3x

-2 x 1/5 = 2/5

-2/3x - 2/5

7 0
4 years ago
Trevor pays $125 in fees each semester plus $650 per college course.
slava [35]
I hope this helps you



for per college course $650


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7 0
3 years ago
Assume that the number of customers who arrive at a water ice stand follows the Poisson distribution with an average rate of 6.4
Nady [450]

Answer:

18.88% probability that three or four customers will arrive during the next 30 minutes

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

Average rate of 6.4 per 30 minutes.

This means that \mu = 6.4

What is the probability that three or four customers will arrive during the next 30 minutes?

P = P(X = 3) + P(X = 4)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 3) = \frac{e^{-6.4}*(6.4)^{3}}{(3)!} = 0.0726

P(X = 4) = \frac{e^{-6.4}*(6.4)^{4}}{(4)!} = 0.1162

P = P(X = 3) + P(X = 4) = 0.0726 + 0.1162 = 0.1888

18.88% probability that three or four customers will arrive during the next 30 minutes

4 0
4 years ago
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