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g100num [7]
3 years ago
6

Jonathan drove 220 miles in four hours. Assuming he drives at the same rate, how far will you travel in seven hours?

Mathematics
1 answer:
Reptile [31]3 years ago
4 0
1)you have to do 220miles multiplie by 7 divide by 4

OR

2)220miles divide by 4 multiplie by7

remaibders is the same thing ,the same answer
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How do you solve what is 30% of $240
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4 years ago
Suppose that the amount of time T a customer spends in a bank is exponentially distributed with an average of 10 minutes. What i
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Answer:

The probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes  is 0.6065.

Step-by-step explanation:

The random variable <em>T</em> is defined as the amount of time a customer spends in a bank.

The random variable <em>T</em> is exponentially distributed.

The probability density function of a an exponential random variable is:

f(x)=\lambda e^{-\lambda x};\ x>0

The average time a customer spends in a bank is <em>β</em> = 10 minutes.

Then the parameter of the distribution is:

\lambda=\frac{1}{\beta}=\frac{1}{10}=0.10

An exponential distribution has a memory-less property, i.e the future probabilities are not affected by any past data.

That is, <em>P</em> (<em>X</em> > <em>s</em> + <em>x</em> | <em>X</em> ><em> s</em>) = <em>P</em> (<em>X</em> > <em>x</em>)

So the probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes  is:

P (X > 15 | X > 10) = P (X > 5)

\int\limits^{\infty}_{5} {f(x)} \, dx =\int\limits^{\infty}_{5}  {\lambda e^{-\lambda x}} \, dx\\=\int\limits^{\infty}_{5}  {0.10 e^{-0.10 x}} \, dx\\=0.10\int\limits^{\infty}_{5}  {e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10}|^{\infty}_{5}\\=[-e^{-0.10 \times \infty}+e^{-0.10 \times 5}]\\=0.6065

Thus, the probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes  is 0.6065.

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