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Brilliant_brown [7]
3 years ago
14

How many solutions does the equation 3(2.4x + 4) = 4.1x + 7 + 3.1x have? Explain.

Mathematics
1 answer:
VMariaS [17]3 years ago
6 0

Answer: there are no solutions

Step-by-step explanation:

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PLEASE HELP ME I WILL GIVE YOU POINTS
patriot [66]
The answer would be C because the y-intercept tells you what y when x is 0. In this case, y is wage and x is years of experience. So, if you have 0 years of experience you get an average of 7.25 dollars per hour, which is C.
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3 years ago
14 to 35 simplest form
Llana [10]
14:35 is a ratio, and of-coarse, as we know, we can transform that into another way of writing a ratio:

14/35

Simplify

14/7 = 2
35/7 = 5

2:5

Answer: 2 to 5
3 0
3 years ago
Read 2 more answers
The city has an average of 13 days of rainfall for April.
zhenek [66]

Using the Poisson distribution, we have that:

  • There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.
  • There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.
  • There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

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

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

The parameters are:

  • x is the number of successes
  • e = 2.71828 is the Euler number
  • \mu is the mean in the given interval.

For this problem, the mean is given as follows:

\mu = 13

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

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

P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

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

P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0

P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003

P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

6 0
2 years ago
Dennis drew the rectangle on grid paper. What is the perimeter of the rectangle dennis drew.
Aleonysh [2.5K]

(2l+2w) is that right

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3 years ago
A box has 6 beads of the same size, but all are different colors. Tania draws a bead randomly from the box, notes its color, and
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It is 1/6 because you only have 6 beads in total meaning that you have a one out of 6 chance of pulling out the yellow bead
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3 years ago
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