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marta [7]
3 years ago
6

Lisa purchased a prepaid phone card for $25 . Long distance calls cost 12 cents a minute using this card. Lisa used her card onl

y once to make a long distance call. If the remaining credit on her card is $19.96 , how many minutes did her call last?
Mathematics
1 answer:
erma4kov [3.2K]3 years ago
8 0
Her call lasted 42 minutes.
Find the difference between 25 and 19.96 and divide the difference by 12.
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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
Gala2k [10]

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.

6 0
3 years ago
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Answer:

35

Step-by-step explanation:

3+4+1=8

120/8=15

3:4:1

45:50:15

50-15=35

please mark as brainliest!!<3

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Answer:

There are no like terms.

7 0
2 years ago
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