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Aleksandr [31]
2 years ago
7

Audra's paper airplane is 13.8 centimeters long. Jordan's paper airplane is 9/10 centimeters shorter than Audra's. How long is J

ordan's paper airplane?
Mathematics
1 answer:
pychu [463]2 years ago
5 0

If Audra's airplane is 13.8 centimeters long, and Jordans airplane is 9/10 centimeters shorter than Audra's. First multiply 13.8 with 9/10, which equals 12.42, next subtract 13.8 and 12.42 which equals 1.38. So Jordan's airplane is I believe 1.38 centimeters long.

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What is 12 2/3 X 2 1/2
vampirchik [111]

Answer:

31.67

Step-by-step explanation:

12 2/3 X 2 1/2

we begin by converting this into improper fractions,

= 38/3 * 5/2

= 95/3

= 31.67

8 0
3 years ago
A rectangular box has a square base. The combined length of a side of the square base, and the height is 20 in. Let x be the len
aniked [119]

Answer:

a. V = (20-x) x^{2} in^{3}  

b . 1185.185 in^{3}

Step-by-step explanation:

Given that:

  • The height:  20  - x (in )
  • Let x be the length of a side of the base of the box (x>0)

a. Write a polynomial function in factored form modeling the volume V of the box.

As we know that, this is a rectangular box has a square base so the Volume of it is:

V = h *x^{2} in^{3}

<=> V = (20-x) x^{2}  in^{3}

b. What is the maximum possible volume of the box?

To  maximum the volume of it, we need to use first derivative of the volume.

<=> dV / Dx = -3x^{2} + 40x

Let dV / Dx = 0, we have:

-3x^{2} + 40x  = 0

<=> x = 40/3

=>the height h = 20/3

So  the maximum possible volume of the box is:

V = 20/3 * 40/3 *40/3

= 1185.185 in^{3}

7 0
2 years ago
A coordinate map of the local grocery store is shown below. Hot dogs are located at the point (6,4) Mustard is located at the po
deff fn [24]

The distance between the 2 locations = 7 units

Step-by-step explanation:

Step 1:

Given,

Location of Hot dogs = (6,4)

Location of Mustard = (6,-3)

We need to find the distance between the two locations

Step 2:

Since the x coordinates of both the location are the same , they lie in the same line.

The first location is 4 units above the x axis and the second location is 3 units below the x axis.

Hence the distance between the 2 locations 4-(-3) = 7 units

Step 3 :

Answer :

The distance between the 2 locations = 7 units

8 0
3 years ago
X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0
1 year ago
Find all roots.<br> 11) x^3 + 6x² + 5x = 0
Dmitrij [34]

Answer:

x=0

x=-1

x=-5

Step-by-step explanation:

To find the roots, we set the equation equal to zero and factor.

x(x²+6x+5)=0           [factor]

x(x+1)(x+5)=0            [set each factor equal to 0]

x=0

x+1=0                       [subtract both sides by 1]

x=-1

x+5=0                      [subtract both sides by 5]

x=-5

Now, we have found three roots, x=0, x=-1, x=-5.

5 0
2 years ago
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