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juin [17]
2 years ago
5

Amare is buying museum souvenirs. He buys 3 animal figures

Mathematics
1 answer:
Juliette [100K]2 years ago
6 0

Answer:

Each animal figure cost $2 dollars and each bookmark cost $4.50 dollars

Step-by-step explanation:

I got this answer is because second part of the question it says "Later he goes back to buy 3 more figures and return 3 of the bookmark spending another $6". The information this gives us is that since he bought nothing but animal figures and bought 3 animal figures and spent $6 dollars that tells us that each animal figure cost $2 dollar knowing this we can figure out the first part since in the first part he spent $33 dollars in total and bought 3 animal figures and 6 bookmarks and we know each animal figure cost $2 dollars we know in total he spent $6 dollar on animal figures and he spent $27 dollar on bookmarks doing division 27 divided by the number of bookmarks he bought we figure out the cost of one bookmark so doing this the answer is one animal figure is $2 dollars and one bookmark is $4.50.

I hope this help this is less of a math problem and more of a logic one this took me a while of rereading to figure out!

Give me a thanks.

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\bf S_{20}=\displaystyle\sum \limits_{n=1}^{\stackrel{\stackrel{n}{\downarrow }}{20}}~\stackrel{\stackrel{a_1}{\downarrow }}{3}(\stackrel{\stackrel{r}{\downarrow }}{1.5})^{n-1}\implies S_{20}=3\left(\cfrac{1-1.5^{20}}{1-1.5} \right)\implies S_{20}=3\left(\cfrac{1-\stackrel{\approx}{3325.3}}{-0.5} \right) \\\\\\ S_{20}=3\left(\cfrac{-3324.3}{-0.5} \right)\implies S_{20}=3(6648.6)\implies S_{20}=19945.8

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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
5^2= (-5)^2 <br> true or false
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It’s True because of 5 square is equal to 25
Than (-5) square is also get the same value as 5^2
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