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qaws [65]
3 years ago
8

5x+2y=7 as a ordered pair

Mathematics
2 answers:
Elenna [48]3 years ago
8 0
(1, 1), 5 + 2 = 7, not that hard!
ICE Princess25 [194]3 years ago
5 0

Step-by-step explanation:

(1, 1), 5 + 2 = 7

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The average monthly rainfall for 6 months was 28.5 mm. If it had rained 1mm more each month what would the average have been? By
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Answer:

29.5 mm

Step-by-step explanation:

We are told that the average monthly rainfall for 6 months was 28.5 mm.

Thus, total for the 6 months = 28.5 × 6 = 171 mm

Now, we are told that it rained 1 mm extra each month.

So extra for the six months = 1 × 6 = 6mm

New total for 6 months = 171 + 6 = 177 mm

So, new average for 6 months = 177/6 = 29.5 mm

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Suppose that the amount of time T a customer spends in a bank is exponentially distributed with an average of 10 minutes. What i
Bad White [126]

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.

8 0
3 years ago
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