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1 year ago

Find the equation of the line y=mx+b form with the slope 3 that passes through the point (5,19)

Mathematics

1 answer:

zaharov [31]1 year ago

6 0

The equation of the line that has a slope of 3 and that passes through the point (5,19) is y = 3x + 4

<h3>Equation of a straight line </h3>

From the question, we are to determine the equation of the line that has a slope of 3 and that passes through the point (5,19)

Using the point-slope form of an equation of a line

y - y₁ = m(x - x₁)

Where m is the slope

and (x₁, y₁) is a point on the line

From the given information

m = 3

x₁ = 5

y₁ = 19

Putting the parameters into the equation, we get

y - 19 = 3(x - 5)

y - 19 = 3x - 15

y = 3x -15 + 19

y = 3x +4

Hence, the equation of the line that has a slope of 3 and that passes through the point (5,19) is y = 3x + 4

Learn more on Equation of a line here: brainly.com/question/16875501

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3 years ago

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3 years ago

Read 2 more answers

assume that when adults with smartphones are randomly selected 15 use them in meetings or classes if 15 adult smartphones are ra

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Answer:

The probability that at least 4 of them use their smartphones is 0.1773.

Step-by-step explanation:

We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.

Also, 15 adult smartphones are randomly selected.

Let X = <em>Number of adults who use their smartphones</em>

The above situation can be represented through the binomial distribution;

$P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; n = 0,1,2,3,.......$

where, n = number of trials (samples) taken = 15 adult smartphones

r = number of success = at least 4

p = probability of success which in our question is the % of adults

who use them in meetings or classes, i.e. 15%.

So, X ~ Binom(n = 15, p = 0.15)

Now, the probability that at least 4 of them use their smartphones is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}$

= $1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})$

= <u>0.1773</u>

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