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r-ruslan [8.4K]

2 years ago

56" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

SCORPION-xisa [38]2 years ago

8 0

Answer:

y ≤-7

Step-by-step explanation:

-8y ≥56

Divide each side by -8. Remember to flip the inequality

-8y/-8 ≤56/-8

y ≤-7

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Suppose the time a child spends waiting at for the bus as a school bus stop is exponentially distributed with mean 7 minutes. De

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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 X represent the time a child spends waiting at for the bus as a school bus stop.

The random variable X 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 X 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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