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3 years ago

The area of a triangle below is 2/5 the base is 6/5 feet what is the length

Mathematics

1 answer:

ddd [48]3 years ago

7 0

Answer:

Step-by-step explanation:

The formula for area of a triangle is 1/2*b*h sq.units

1/2*6/5*h=2/5

3/5*h=2/5

h=2/5*5/3=2/3 feet

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Given f(x) is a linear function with slope 3 and y-intercept -5, what is the equation of f(x) in standard form?

lawyer [7]

The standard form:

$Ax+By=C$

The slope-intercept form:

$y=mx+b$

m - slope

b - y-intercept

We have m = 3 and b = -5.

Substitute:

$y=3x-5$ subtract 3x from both sides

$-3x+y=-5$ change the signs

$3x-y=5$

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3 years ago

Write 10 1/4 as a decimal number

EastWind [94]

Answer:

10.25

Step-by-step explanation:

1/4- is 0.25

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3 years ago

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What is the equation in point-slope form if a line that passes through points (3,-5) and (-8,4)?

nignag [31]

The slope-point formula:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have (3, -5) and (-8, 4).

Substitute:

$m=\dfrac{4-(-5)}{-8-3}=\dfrac{9}{-11}=-\dfrac{9}{11}\\\\y-4=-\dfrac{9}{11}(x-(-8))\\\\y-4=-\dfrac{9}{11}(x+8)\to C)$

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3 years ago

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$

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3 years ago