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Juli2301 [7.4K]
3 years ago
8

Find the value of x. When a leg is 6 units and the hypotenuse is 12 units.

Mathematics
1 answer:
DIA [1.3K]3 years ago
3 0

Answer:

6√3

Step-by-step explanation:

6^2+x^2=12^2

36+x^2=144

-36          -36

x^2=108

x=6√3

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Kristian ordered 3 DVDS by mail each dvd cost the same amount with a $6 shipping charge the total cost was $42 how much did each
Natasha2012 [34]

Answer:

$12

Step-by-step explanation:

42 - 6 = 36

36/3

12

7 0
3 years ago
Solve for x<br> X^2+6x+9=20
Masja [62]

Answer:

x = 1.47  or x = -7.47

Step-by-step explanation:

x²+6x+9=20

This is a quadratic equation

x²+6x+9-20=0

x²+6x-11=0

Step 1 : Write the quadratic formula

x = <u>-b±√b²-4(a)(c)</u>

              2a

Step 2 : Substitute values in the formula

a = 1

b = 6

c = -11

x = <u>-6±√6²-4(1)(-11)</u>

              2(1)

x = <u>-6±√80</u>

          2

x = -3 + 2√5 or x = -3 - 2√5

x = 1.47         or x = -7.47

!!

4 0
3 years ago
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Which of the following points lies in Quadrant II?
RUDIKE [14]
The answer is d. (-3,8.)

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The functions q and r are defined as follows.
UkoKoshka [18]
Q(3) = -2(3)+2 = -4
r(q(3)) = (-4)^2-1 = 15
3 0
3 years ago
When an automobile is stopped with a roving safety patrol,each tire is checked for tire wear, and each headlight is checkedto se
ryzh [129]

Answer:

a) Joint ptobability distribution

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

P(x,y)=P_x(x)*P_y(y)

Then, the joint probabilities are:

X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025

X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06

X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01

X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56

We can calculate P(X<= 1) * P(Y<=1)

P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56

Both calculations give the same result.

c) Probability of no violations

P(X+Y=0)=P(0,0)=0.3

d) P(X + Y <= 1)

P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53

5 0
3 years ago
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