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ANEK [815]
3 years ago
10

A card was selected at random from a standard deck of cards. The suit of the card was recorded, and then the card was put back i

n the deck. The table shows the results after 40 trials.
What is the relative frequency of selecting a club?


17%

20%

25%

30%
Outcome Club Diamond Heart Spade
Number of trials 8 12 11 9
Mathematics
2 answers:
scoundrel [369]3 years ago
6 0

Answer:

30%

Step-by-step explanation:

Gennadij [26K]3 years ago
4 0

Answer:

30%

Step-by-step explanation:

Relative frequency : total number of diamonds / total number of cards drawn = 11/40=0.3

The answer is 30%

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8) through: (-3, -2), perp. to y = x – 1<br> A) y=-5x – 1 B) y=-4x – 5<br> C) y=-x – 5 D) y=-5x – 4
nexus9112 [7]

<u>Answer:</u>

The equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

<u>Solution:</u>

Given, line equation is y = x – 1 ⇒ x – y – 1 = 0. And a point is (-3, -2)

We have to find the line equation which is perpendicular to above given line and passing through the given point.

Now, let us find the slope of the given line equation.

\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-1}=1

We know that, <em>product of slopes of perpendicular lines is -1. </em>

So, 1 \times slope of perpendicular line =  -1

slope of perpendicular line = -1

Now let us write point slope form for our required line.

\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)

y – (-2) = -1(x – (-3))

y + 2 = -1(x + 3)

y + 2 = -x – 3

x + y + 2 + 3 = 0

x + y + 5 = 0

y = -x -5

Hence the equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

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4 years ago
Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)
vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

x^{\frac{2}{3}}+y^{\frac{2}{3}}=4

We have to find the equation of tangent line to the given curve at point (-3\sqrt3,1)

By using implicit differentiation, differentiate w.r.t x

\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0

Using formula :\frac{dx^n}{dx}=nx^{n-1}

\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}

\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}

\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}

Substitute the value x=-3\sqrt3,y=1

Then, we get

\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}

\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}

Slope of tangent=m=\frac{1}{\sqrt3}

Equation of tangent line with slope m and passing through the point (x_1,y_1) is given by

y-y_1=m(x-x_1)

Substitute the values then we get

The equation of tangent line is given by

y-1=\frac{1}{\sqrt3}(x+3\sqrt3)

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y=\frac{x}{\sqrt3}+3+1

y=\frac{x}{\sqrt3}+4

This is required equation of tangent line to the given curve at given point.

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