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coldgirl [10]
3 years ago
12

Please help, it would be much appreciated.A certain state uses the following progressive

Mathematics
1 answer:
liraira [26]3 years ago
6 0

Answer:

$3,300

Step-by-step explanation:

$60,000 falls into the $50,001 - $100,000 range, so 5.5% is owed on it.

5.5% → 0.055

60,000 * 0.055 = $3,300

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How do u put 72 7/10 into decimal?
NNADVOKAT [17]

Answer:

72.7

Step-by-step explanation:

First multiply 72 by 10

because 72 is 72 10s, making it a whole number.

then add 720 (72 times 10) to 7.

then all of that over 10.

then just divide.

727/10

72.7

this is what you do for all mixed numbers like that one.

7 0
3 years ago
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HELP ASAP
galben [10]

I don’t have a very detailed explanation, however the answer should be “A 90-degree counterclockwise rotation about the origin followed by a translation 1 unit to the left”.

You can clearly see a 90° angle here, it’s obvious. Also, the polygon switch counterclockwise. so, you can see.

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3 years ago
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doughnut are sold in bag and cartons. a bag holds 4 doughnuts and a carton holds 10 doughnuts. tom buys b bags of doughnuts and
Radda [10]
As there are 10 doughnuts in each cartoon and 4 in each bag, the formula would be:

t=10c+4b
3 0
4 years ago
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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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Mkey [24]

Answer:

-11

Step-by-step explanation:

5 0
3 years ago
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