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4 years ago

Graph the equations to solve the system Y equals 2x+7 and 1/2y=x+7/2

Mathematics

1 answer:

FinnZ [79.3K]4 years ago

3 0

Given equation is y=2x+7 and $\frac{1}{2}y=x+\frac{7}{2}$

Let's simplify the 2nd equation $\frac{1}{2}y=x+\frac{7}{2}$ before we can start graph so that calculation will be easy

$\frac{1}{2}y=x+\frac{7}{2}$

multiply both sides by 2 to cancel out fractions

$2*\frac{1}{2}y=2*x+2*\frac{7}{2}$

y=2x+7

which is exactly same as the first equatoin so graph of both will be exactly same and solution will be infinitely many solutions.

y=2x+7 has y-intercept 7 so first point will be (0,7). Slope is 2 so rise 2 unit up then 1 right and graph the new point.

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3 years ago

Find the value 'X' in this given picture

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Hi.
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4 years ago

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vredina [299]

Using the binomial distribution, it is found that there is a 0.056 = 5.6% probability that more than 7 will make a purchase.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

There are 12 customers, hence n = 12.

The probability of any of them making a purchase is of p = 0.4.

The probability that more than 7 will make a purchase is given by:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12).

Hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.4)^{8}.(0.6)^{4} = 0.042$

$P(X = 9) = C_{12,9}.(0.4)^{9}.(0.6)^{3} = 0.012$

$P(X = 10) = C_{12,10}.(0.4)^{10}.(0.6)^{2} = 0.02$

$P(X = 11) = C_{12,11}.(0.4)^{11}.(0.6)^{1} \approx 0$

$P(X = 12) = C_{12,12}.(0.4)^{12}.(0.6)^{0} \approx 0$

Then:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.042 + 0.012 + 0.02 + 0 + 0 = 0.056.

0.056 = 5.6% probability that more than 7 will make a purchase.

More can be learned about the binomial distribution at brainly.com/question/24863377

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3 years ago

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3 years ago

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