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2 years ago

10

You are making a poster that will have a uniform border, as shown. the total area of the poster is 722 square inches. find the w

idth of the border to the nearest inch

1 answer:

olga55 [171]2 years ago

6 0

I FOUND YOUR COMPLETE QUESTION IN OTHER SOURCES.
PLEASE SEE ATTACHED IMAGE.
For this case the area is given by:
A = (22 + x) * (28 + x) = 722
Rewriting we have:
616 + 22x + 28x + x ^ 2 = 722
x ^ 2 + 50x + 616 - 722 = 0
x ^ 2 + 50x - 106 = 0
Solving the polynomial we have:
x1 = 2.04
x2 = -52.04
We take the positive root:
x = 2.04 inches:
Answer:
The width of the border to the nearest inch is:
x = 2inches

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The dot plot and box plot show different displays of the ages of runners on a track team. Which statements best describe the dot

kirill115 [55]

Answer:

I think is is a bit b

Step-by-step explanation:

a or b

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2 years ago

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

UNO [17]

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600$

Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

8 0

2 years ago

Find the equation of the plane through the point (2,5,7) that is parallel to the line r=(3i+2j−2k)+t(i+2j+9k) and perpendicular

insens350 [35]

The plane we want to find has general equation

$a(x-2)+b(y-5)+c(z-7)=0$

with $a,b,c$ not equal to 0, and has normal vector

$\vec n=a\,\vec\imath+b\,\vec\jmath+c\,\vec k$

$\vec n$ is perpendicular to both the normal vector of the other plane, which is $4\,\vec\imath+5\,\vec\jmath+6\,\vec k$ , as well as the tangent vector to the line $\vec r(t)$ , which is $\vec\imath+2\,\vec\jmath+9\,\vec k$ .

This means the dot product of $\vec n$ with either vector is 0, giving us

$\begin{cases}4a+5b+6c=0\\a+2b+9c=0\end{cases}$

Suppose we fix $c=1$ . Then the system reduces to

$\begin{cases}4a+5b=-6\\a+2b=-9\end{cases}$

and we get

$(4a+5b)-4(a+2b)=-6-4(-9)\implies-3b=30\implies b=-10$

$a+2(-10)=-9\implies a=11$

Then one equation for the plane could be

$\boxed{11(x-2)-10(y-5)+(z-7)=0}$

or in standard form,

$\boxed{11x-10y+z=-21}$

The solution is unique up to non-zero scalar multiplication, which is to say that any equation $(11x-10y+z)k=-21k$ would be a valid answer. For example, suppose we instead let $c=2$ ; then we would have found $a=22$ and $b=-20$ , but clearly dividing both sides of the equation

$22(x-2)-20(y-5)+2(z-7)=0$

by 2 gives the same equation as before.

7 0

2 years ago

The volume of air inside a rubber ball with radius r can be found using the function V(r) = four-thirds pi r cubed. What does V

aleksklad [387]

Answer:

The answer is explained below

Step-by-step explanation:

Given that The volume of air inside a rubber ball with radius r can be found using the function V(r) = $\frac{4}{3}\pi r^3$ , this means that the volume of the air inside the rubber ball is a function of the radius of the rubber ball, that is as the radius of the rubber ball changes, also the volume of the ball changes.

As seen from the function, the radius is directly proportional to the volume of the ball, if the radius increases, the volume also increases.

$V(\frac{5}{7} )$ is equal to the volume of the ball when the radius of the ball is $\frac{5}{7}$ . Therefore:

$V(\frac{5}{7} )=\frac{4}{3} \pi *(\frac{5}{7} )^3=1.53\ unit^3$

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2 years ago

The value -2 is a lower bound for the zeros of the function shown below.

Juliette [100K]

Answer:

a) true

Step-by-step explanation:

The bottom-line numbers from synthetic division <em>alternate signs</em>, indicating that -2 is a lower bound. The given statement is true.

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If the signs were all positive, it would indicate the proposed zero is an <em>upper bound</em>.

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A graph shows all real zeros are greater than -2.

4 0

2 years ago