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choli [55]
3 years ago
12

5 and 1/4 - 3 and 3/4

Mathematics
2 answers:
slavikrds [6]3 years ago
6 0
Hi there.

5.25 - 3.75 (I converted the fractions to decimals to make things a little more simple, but I'll convert them back to fractions.)

First, subtract .25 from 5.25 and 3.75. This allows the process to be easier to break down without making mistakes.

Now, subtract .50 from 5 and 3.50.
We now have 4.50 - 3.00; simple enough from here, right?

4.50 - 3.00 = 1.50, but if we convert it back, we have 1 and 1/2.

Your answer is 1 and 1/2 or 1.50

I hope this helps!
s2008m [1.1K]3 years ago
4 0
The answer is 6/4
Convert the mixed number <span><span>5 <span>1/4</span></span></span> into an improper fraction first by multiplying the denominator <span><span>(4)</span></span> by the whole number part <span><span>(5)</span></span> and add the numerator <span><span>(1)</span></span> to get the new numerator. Place the new numerator <span><span>(21)</span></span> over the old denominator <span><span>(4)</span></span><span><span><span>21/4</span>−3<span>3/4</span></span></span>Convert the mixed number <span><span>3 <span>3/4</span></span></span> into an improper fraction first by multiplying the denominator <span><span>(4</span><span>)</span></span> by the whole number part <span><span>(3)</span></span> and add the numerator <span><span>(3)</span></span> to get the new numerator. Place the new numerator <span><span>(15)</span></span> over the old denominator <span><span>(4)</span></span>.<span><span><span>21/4</span>−<span>15/4</span></span></span>Combine the numerators over the common denominator.<span><span><span>21−1⋅15/</span>4</span></span>Multiply <span><span>−1</span></span> by <span>15</span> to get <span><span>−15</span></span>.<span><span><span>21−15/</span>4</span></span>Subtract <span>15</span> from <span>21</span> to get <span>6</span>.<span>6/4</span>
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<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)

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