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yuradex [85]
3 years ago
15

What does the line of -4 x + 3y = 6 look like

Mathematics
2 answers:
Llana [10]3 years ago
8 0

Answer:

The line slopes upwards from left to right wit a positive gradient and cuts the y-axis at y=2 and the x-axis at x=-3/2

Step-by-step explanation:

We first rearrange the equation to the order y=mx+c where m is the gradient and c the y intercept.

3y=4x+6

y=(4/3)x+2

The gradient is therefore 4/3 and the y intercept is 2.

At the c intercept, y=0

0=(4/3)x +2

(4/3)x=-2

x=-2×3/4

=3/2

The line slopes upwards from left to right with a positive gradient and cuts the y-axis at y=2 and the x-axis at x=-3/2

dezoksy [38]3 years ago
4 0
The answer is y=4/3x+2. The explanation is shown in the picture.
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The ratio of Boys to Girls at Sweet water Middle School is 5:4. If there are 50 boys on an academic team, how many girls are the
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4 years ago
Cars arrive at the Wendy's drive-through at a rate of 1 car every 5 minutes between the hours of 11:00 PM and 1:00 AM. on Saturd
mestny [16]

Answer:

1) P(X = 8) = 0.1033

P(X = 9) = 0.0688

2) Expected number of 200 restaurants in which exactly 8 customers use the drive-through: 20.66

Expected number of 200 restaurants in which exactly 9 customers use the drive-through: 13.76

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 time interval.

Question 1. Use the Poisson distribution to calculate the probability that exactly 8 cars will use the drive-through between 12:00 midnight and 12:30 AM on a Saturday night at Wendy's. Do the same for exactly 9 cars.

Cars arrive at the Wendy's drive-through at a rate of 1 car every 5 minutes between the hours of 11:00 PM and 1:00 AM. on Saturday nights. This means that during 30 minutes, 6 cars expected to arrive. So \mu = 6.

P(X = 8)

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

P(X = 8) = \frac{e^{-6}*(6)^{8}}{(8)!} = 0.1033

P(X = 9)

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

P(X = 9) = \frac{e^{-6}*(6)^{9}}{(9)!} = 0.0688

Question 2. At how many of the 200 restaurants in the survey would you expect exactly 8 customers to use the drive-through? exactly 9 customers?

There is a 10.33 probability that 8 customers would use the drive through for each restaurant.

So of 200, the expected number is

E(X) = 200*0.1033 = 20.66

There is a 6.88 probability that 9 customers would use the drive through for each restaurant.

So of 200, the expected number is

E(X) = 200*0.0688 = 13.76

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