1/3x^2 by -2/9x divide it

A track star runs two races on a certain day. The probability thathe wins the first race is 0.7, the probability that he wins th

NARA [144]

Answer:

a) 80% probability that he wins at least one race.

b) 30% probability that he wins exactly one race.

c) 20% probability that he wins neither race.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that he wins the first race.

B is the probability that he wins the second race.

C is the probability that he does not win any of these races.

We have that:

$A = a + (A \cap B)$

In which a is the probability that he wins the first race but not the second and $A \cap B$ is the probability that he wins both these races.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that he wins both races is 0.5.

This means that $A \cap B = 0.5$

The probability that he wins the second race is 0.6

This means that $B = 0.6$

$B = b + (A \cap B)$

$0.6 = b + 0.5$

$b = 0.1$

The probability that he wins the first race is 0.7.

This means that $A = 0.6$

$A = a + (A \cap B)$

$0.7 = a + 0.5$

$a = 0.2$

A) he wins at least one race.

This is

$P = a + b + (A \cap B) = 0.2 + 0.1 + 0.5 = 0.8$

There is an 80% probability that he wins at least one race.

B) he wins exactly one race.

This is

$P = a + b = 0.2 + 0.1 = 0.3$

There is a 30% probability that he wins exactly one race.

C) he wins neither race

Either he wins at least one race, or he wins neither. The sum of these probabilities is 100%.

From a), we have that there is an 80% probability that he wins at least one race.

So there is a 100-80 = 20% probability that he wins neither race.

6 0

3 years ago

A rectangular park of length 60 m breadth 50 m encloses w volleyball court of length 18 m and breadth 10 m. Find the area of the

Zanzabum

Given

A rectangular park of length 60 m breadth 50 m encloses with volleyball court of length 18 m and breadth 10 m.

To find:

The area of the park excluding the court at the rate of Rs 110 per square meter.

Solution:

Area of a rectangle is:

$Area=length \times breadth$

Area of whole park is:

$A_1=60 \times 50$

$A_1=3000$

Area of volleyball court is:

$A_2=18 \times 10$

$A_2=180$

Now, the area of the park excluding the court is:

$A=A_1-A_2$

$A=3000-180$

$A=2820$

Therefore, the area of the park excluding the court is 2820 square meter.

4 0

3 years ago

In parallelogram WXYZ, what is CY? CY=___ ft

dangina [55]

The correct answer is 11 ft.

Explanation:
The diagonals of a parallelogram bisect each other. This means that the portion marked x+4 is the same length as the portion marked 2x-7; this gives us the equation x+4=2x-7.

Subtract x from each side:
x+4-x=2x-7-x
4=x-7.

Add 7 to both sides:
4+7=x-7+7
11=x

6 0

3 years ago

Read 2 more answers

X ſocm Jicm A) 2V5 cm B) V13 cm D) V19 cm C) 1 cm ASAP

bulgar [2K]

Answer:

using pythagoras theorem

hyp 2=opp 2+adj 2

=( √6)(√6)+(√7)(√7)

hyp 2=13

hyp=√13

3 0

3 years ago

Read 2 more answers

For which system of equations is (5, 3) the solution? A. 3x – 2y = 9 3x + 2y = 14 B. x – y = –2 4x – 3y = 11 C. –2x – y = –13 x

Alla [95]

The correct answer is:

D) $\left \{ {{2x-y=7} \atop {2x+7y=31}} \right.$ .

Explanation:

We solve each system to find the correct answer.

For A:
$\left \{ {{3x-2y=9} \atop {3x+2y=14}} \right.$

Since we have the coefficients of both variables the same, we will use elimination to solve this.

Since the coefficients of y are -2 and 2, we can add the equations to solve, since -2+2=0 and cancels the y variable:
$\left \{ {{3x-2y=9} \atop {+(3x+2y=14)}} \right. 
\\
\\6x=23$

Next we divide both sides by 6:
6x/6 = 23/6
x = 23/6

This is not the x-coordinate of the answer we are looking for, so A is not correct.

For B:
$\left \{ {{x-y=-2} \atop {4x-3y=11}} \right.$

For this equation, it will be easier to isolate a variable and use substitution, since the coefficient of both x and y in the first equation is 1:
x-y=-2

Add y to both sides:
x-y+y=-2+y
x=-2+y

We now substitute this in place of x in the second equation:
4x-3y=11
4(-2+y)-3y=11

Using the distributive property, we have:
4(-2)+4(y)-3y=11
-8+4y-3y=11

Combining like terms, we have:
-8+y=11

Add 8 to each side:
-8+y+8=11+8
y=19

This is not the y-coordinate of the answer we're looking for, so B is not correct.

For C:
Since the coefficient of x in the second equation is 1, we will use substitution again.

x+2y=-11

To isolate x, subtract 2y from each side:
x+2y-2y=-11-2y
x=-11-2y

Now substitute this in place of x in the first equation:
-2x-y=-13
-2(-11-2y)-y=-13

Using the distributive property, we have:
-2(-11)-2(-2y)-y=-13
22+4y-y=-13

Combining like terms:
22+3y=-13

Subtract 22 from each side:
22+3y-22=-13-22
3y=-35

Divide both sides by 3:
3y/3 = -35/3
y = -35/3

This is not the y-coordinate of the answer we're looking for, so C is not correct.

For D:
Since the coefficients of x are the same in each equation, we will use elimination. We have 2x in each equation; to eliminate this, we will subtract, since 2x-2x=0:

$\left \{ {{2x-y=7} \atop {-(2x+7y=31)}} \right. 
\\
\\-8y=-24$

Divide both sides by -8:
-8y/-8 = -24/-8
y=3

The y-coordinate is correct; next we check the x-coordinate Substitute the value for y into the first equation:
2x-y=7
2x-3=7

Add 3 to each side:
2x-3+3=7+3
2x=10

Divide each side by 2:
2x/2=10/2
x=5

This gives us the x- and y-coordinate we need, so D is the correct answer.

7 0

3 years ago

1/3x^2 by -2/9x divide it ​

1/3x^2 by -2/9x divide it