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Advocard [28]
3 years ago
15

Megan and Rosa are throwing sandbags during a workout. Megan's farthest throw measured 8.9 meters. Rosa's farthest throw measure

d 10.05 meters. How much further did Rosa throw the sandbag than Meghan?
Mathematics
1 answer:
ankoles [38]3 years ago
6 0
10.05-8.9=1.15. Rosa threw it 1.15 meters farther than Meghan
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Which term means 1000?
Shalnov [3]

Answer:

K

Step-by-step explanation:

The term "K" can be used in place of thousand, such as 5k

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~theLocoCoco

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it took Simon 33 minutes to run 5.5 miles. did he run faster or slower than 1 mile every 5 minutes how can you tell ​
vitfil [10]
Simon ran slower than 1 mile every 5 minutes.

Because 33 / 5.5 = 6. So he ran 1 mile every 6 minutes, not 5.
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Rectangular pyramid:volume 448 inches cubed, base edge 12 inches.;base length 8 inches
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Whats your question?
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3 years ago
A triangle has side lengths measuring 2x + 2 ft x + 3 ft, and
Marta_Voda [28]

Answer:

3x+5 < n < x-1

Step-by-step explanation:

we know that

The <u><em>Triangle Inequality Theorem</em></u> states that the sum of any 2 sides of a triangle must be greater than the measure of the third side

so

Applying the Triangle Inequality Theorem

1) (2x+2)+(x+3) > n

solve for n

3x+5> n

Rewrite

n

2) (x+3)+n> 2x+2

n> x-1

therefore

3x+5 < n < x-1

5 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
2 years ago
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