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3 years ago

(4,7) y=3x+6 write an equation passing through the point and parallel to the given line

Mathematics

1 answer:

Elis [28]3 years ago

4 0

So, we will be using point-slope form to construct this equation.

Point-slope form's general formula is as follows:

$Y-Y_1=m(X-X_1)$

M is the slope,

Y₁ is the Y-coordinate of the point

X₁ is the X- coordinate of the point.

So, our slope is 3 (since the equation is given in slope-intercept form)

Our Y₁ is 7 and out X₁ is 4

So, now we just plug our numbers into the equation.

$Y-7 = 3(X-4)$

There's your equation.

Hope this helps!

You might be interested in

The diagram shows a 9cm x 7cm rectangle-based pyramid. all the diagonal sides - TA, TB, TC, AND TD- are length 12cm. M is the mi

lakkis [162]

Answer:

∠TAC is approximately equal to 61.6°

Step-by-step explanation:

The given parameters for the pyramid are;

The dimension for the rectangular base are; Length = 9 cm, width = 7 cm

The length of the diagonal sides, TA, TB, TC, and TD = 12 cm each

The midpoint of the rectangular base = Point M

The diagonal AC = AM + MC

AM = MC as given M is the midpoint of the rectangular base

∴ AC = AM + MC = 2·AM

By Pythagoras' theorem, AC = √(9² + 7²) = √130

AC = √130 cm

∴ AM = AC/2 = (√130)/2 cm

Alternatively, AM = √((9/2 cm)² + (7/2 cm)²) = √(32.5) cm

∠TAC = ∠TAM

By trigonometric ratios, we have;

$cos (\theta) = \dfrac{Length \ of \ adjacent \ side \ to \ angle }{Length \ of \ hypotenuse\ side \ to \ angle}$

$\therefore cos (\angle TAM) = cos (\angle TAC) = \dfrac{\left (\dfrac{\sqrt{130} }{2} \right )}{12} = \dfrac{\sqrt{130} }{2 \times 12} = \dfrac{\sqrt{130} }{24}$

$\angle TAC = arccos \left ( \dfrac{\sqrt{130} }{24} \right ) \approx 61.6 ^{\circ} \ to 1 \ decimal \ place$

6 0

2 years ago

Tom buys a shirt for 22.00 plus 7% sales tax. His cousin buys a shirt for 18.00 plus sales tax. How much do tom and his cousin p

Novay_Z [31]

Hey You!

Tom:

22*7% = 1.54

His Cousin:

18*7% = 1.26

22 + 1.54 = 23.54

18 + 1.26 = 19.26

So, Tom pays $23.54 and His Cousin pays $19.26 and Tom payed more tax.

5 0

3 years ago

Read 2 more answers

(43 points) In the US, 85% of the population has Rh positive blood. Suppose we take a random sample of 6 persons and let Y denot

VladimirAG [237]

Answer:

a) Binomial distribution with parameters p=0.85 q=0.15 n=6

b) 62.29%

c) 2.38%

d) See explanation below

Step-by-step explanation:

We could model this situation with a binomial distribution

$P(6;k)=\binom{6}{k}p^kq^{6-k}$

where P(6;k) is the probability of finding exactly k people out of 6 with Rh positive, p is the probability of finding one person with Rh positive and q=(1-p) the probability of finding a person with no Rh.

$\bf P(Y=k)=\binom{6}{k}(0.85)^k(0.15)^{6-k}$

The probability that Y is less than 6 is

P(Y=0)+P(Y=1)+...+P(Y=5)

Let's compute each of these terms

$P(Y=0)=P(6;0)=\binom{6}{0}(0.85)^0(0.15)^{6}=1.139*10^{-5}$

$P(Y=1)=P(6;1)=\binom{6}{1}(0.85)^1(0.15)^{5}=0.0000387281$

$P(Y=2)=P(6;2)=\binom{6}{2}(0.85)^2(0.15)^{4}=0.005486484$

$P(Y=3)=P(6;3)=\binom{6}{3}(0.85)^3(0.15)^{3}=0.041453438$

$P(Y=4)=P(6;4)=\binom{6}{4}(0.85)^4(0.15)^{2}=0.176177109$

$P(Y=5)=P(6;5)=\binom{6}{5}(0.85)^5(0.15)^{1}=0.399334781$

and adding up these values we have that the probability that Y is less than 6 is

$\sum_{i=1}^{5}P(Y=i)=0.622850484\approx 0.6229=62.29\%$

In this case is a binomial distribution with n=200 instead of 6.

p and q remain the same.

The mean of this sample would be 85% of 200 = 170.

In a binomial distribution, the standard deviation is

$s = \sqrt{npq}$

In this case

$\sqrt{200(0.85)(0.15)}=5.05$

<em>Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05</em>

So, the approximate probability that there are fewer than 160 persons with Rh positive blood in a sample of 200 would be the area under the normal curve to the left of 160

(see picture attached)

We can compute that area with a computer and find it is

0.0238 or 2.38%

d)<em> In order to approximate a binomial distribution with a normal distribution we need a large sample like the one taken in c).</em>

In general, we can do this if the sample of size n the following inequalities hold:

$np\geq 5 \;and\;nq \geq 5$