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Ratling [72]
2 years ago
6

9. If H is the midpoint of GI, find GH. 5x + 2 9x - 10 FIND GH I inserted the graph

Mathematics
1 answer:
sertanlavr [38]2 years ago
7 0

Answer:

GH = 17

Step-by-step explanation:

Given that H is the midpoint of segment GI, and GH = 5x + 2, HI = 9x - 10, GH is congruent to HI. This implies GH = HI.

Therefore:

GH = HI, which will give us the following equation:

5x + 2 =  9x - 10

Solve for x

5x + 2 - 9x =  9x - 10 - 9x (subtracting 9x from each side)

-4x + 2 = -10

-4x + 2 - 2 = -10 - 2 (subtracting 2 from each side)

-4x = -12

\frac{-4x}{-4} = \frac{-12}{-4} (dividing both sides by -4)

x = 3

GH = 5x + 2

Plug in the value of x

GH = 5(3) + 2 = 15 + 2

GH = 17

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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
2 years ago
The Sweet water High School Project Graduation committee is hosting a dinner-and-dance fundraiser at the Sweet water Community C
yan [13]

Answer:

<h2> 105 tickets</h2>

Step-by-step explanation:

To solve this problem we need to model an equation to represent the situation first.

the goal is to archive $7500 in the even, bearing in mind that there is a cost of $375 fee for rent, we need to put this amount into consideration

let the number of tickets be x

so

75x-375>=7500--------1

Equation 1 above is a good model for the equation

we can now solve for x to determine the number of tickets to be sold to archive the aim

75x-375>=7500--------1

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4 0
2 years ago
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A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that t
Alina [70]

Answer:

75.76% probability that there will be 10 or more customers at this bank in one hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

A bank gets an average of 12 customers per hour.

This means that \mu = 12

Find the probability that there will be 10 or more customers at this bank in one hour.

Either there are less than 10 customers, or there are 10 or more. The sum of the probabilities of these events is 1. Then

P(X < 10) + P(X \geq 10) = 1

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P(X \geq 10) = 1 - P(X < 10)

In which

P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

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P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

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Then

P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576

75.76% probability that there will be 10 or more customers at this bank in one hour.

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2 years ago
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2 years ago
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