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denis-greek [22]
3 years ago
14

a store is offering a 20% discount on all sales over $50. If you purchase a shirt and a pair of leans for $62.50, what is the am

ount of the discount you would receive?
Mathematics
1 answer:
vodomira [7]3 years ago
7 0

Answer:

you would receive 12.5% of the discount

Step-by-step explanation:

20% off of 62.50 is 50; so you would pay $50

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During a period of 45 days, Delynn played soccer on 3/5 of the days. On how many days did she play soccer?
Archy [21]

Answer:

Step-by-step explanation:

No. of days Delynn played soccer = 3/5 of 45

              = (3/5) * 45   = 3*45/5 = 3*9

              = 27 days

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olga nikolaevna [1]

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the answer is:

j= 76

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What is 0.1265625 rounded to the nearest tenth
katrin2010 [14]

Answer:

0.1 is 0.1265525 rounded to the nearest tenth

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Katyanochek1 [597]

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6

Step-by-step explanation:

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8 0
2 years ago
Let U1, ..., Un be i.i.d. Unif(0, 1), and X = max(U1, ..., Un). What is the PDF of X? What is EX? Hint: Find the CDF of X first,
Kryger [21]

Answer:

E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}

Step-by-step explanation:

A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".

We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).

If we select a value x \in (0,1) we want this:

max(U_1, ....,U_n) \leq x

And we can express this like that:

u_i \leq x for each possible i

We assume that the random variable u_i are independent and P)U_i \leq x) =x from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:

P(X \leq x) = P(U_1 \leq 1, ...., U_n \leq x) \prod P(U_i \leq x) =\prod x = x^n

And then cumulative distribution would be expressed like this:

0, x \leq 0

x^n, x \in (0,1)

1, x \geq 1

For each value x\in (0,1) we can find the dendity function like this:

f_X (x) = \frac{d}{dx} F_X (x) = nx^{n-1}

So then we have the pdf defined, and given by:

f_X (x) = n x^{n-1} , x \in (0,1)  and 0 for other case

And now we can find the expected value for the random variable X like this:

E(X) =\int_{0}^1 s f_X (x) dx = \int_{0}^1 x n x^{n-1}

E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}

6 0
3 years ago
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