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Svet_ta [14]
3 years ago
12

Of the 48 oranges in a carton, one-third of them were rotten. Ben and David took the remaining good oranges. If Ben took 8 more

oranges than David, how many oranges did David take?
Mathematics
2 answers:
vekshin13 years ago
7 0
<span>First you need to find out what is 1/3 of 48 which is 16 
Then 48-16 which is 32
so now you know there are 32 oranges left 
</span>David took 12 oranges and Ben took 20 oranges
12 + 20 = 32 there you go!!
kow [346]3 years ago
3 0
First you need to find out what is 1/3 of 48 which is 16 
Then 48-16 which is 32
so now you know there are 32 oranges left 
Ben took 8 more then David what do you think answer would be??
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Step-by-step explanation:

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Answer:

The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.

The random variable <em>X</em> is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is,\lambda=\frac{1}{\mu}=\frac{1}{7}.

The probability density function of <em>X</em> is:

f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx

                      =\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148

Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

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