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anygoal [31]
3 years ago
14

Sue scored a total of 35 points in two games. She scored 6 times as many points in the second game as in the first. How many mor

e points did she score in the second game?
Mathematics
1 answer:
Nata [24]3 years ago
7 0

x = points in first game

y = points in second game

6x=y   to make them equal multiply the first games points by 6

x+y=35    the points in both games

x +6x = 35   substitute for y  (y=6x)

7x=35   combine like terms

x=5

y=6x

y=6(5)

y=30 points

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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
before a renovation, a movie theater had 149 seats. After the renovation, the theater had 180 seats. what is the approximate per
liberstina [14]

180 - 149 = 31

x/100 *149 = 31

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6 * 6 = 36

We add up all of the numbers:

48 + 36 = 84

84 in^2 is the answer. 

Hope this helps!
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Answer:

y = -2/7x + 4

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