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Jlenok [28]
4 years ago
9

What is 48÷1261 with the remainder

Mathematics
1 answer:
Agata [3.3K]4 years ago
3 0

Hi Evangrayson8,

To get the answer to this we need to divide. So, 48 ÷ 1261 would be 0.03806502775.

Hope this helps

Cupkake~

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Please help!! 30 points!!
nasty-shy [4]

Answer:

The expected sum is $220 and the standard deviation is $65

Step-by-step explanation:

In order to get the expected sum you add the two expected sums you are given

To get the standard deviation you add the two given standard deviations

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4 years ago
Let f (x)=x-2 and g(x)=x to the second power -7x-9.find f (g(-1))
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f(g(-1)) = - 3

Evaluate g(-1) and substitute into f(x)

g(-1) = (-1)² -7(-1) - 9 = 1 + 7 - 9 = - 1

f(g(-1)) = (-1) - 2 = - 3


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3 years ago
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tino4ka555 [31]
The answer is-B (x-9) (x-9)
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A 4-digit PIN number is selected. What it the probability that there are no repeated digits?
geniusboy [140]

Answer:

P(A)=0.504 or 50.4%

Step-by-step explanation:

The <em>probabilities</em> constitute a branch of mathematics that deals with measuring or quantitatively determining the possibility that an event or experiment produces a certain result. The probability is based on the study of combinatorics and is the necessary foundation of statistics.

A <em>permutation</em> is an ordered arrangement of objects in a group, without repetitions.

A 4-digit PIN is selected. What is the probability that there are no repeated digits?

First, there are only 10 possible values for each digit of the PIN (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). The total possibles 4 PINs are 10^{4} =10000

There's no repeated numbers and all the four numbers have to be different, which means we have to select 4 numbers out of 10 numbers. Using the permutation equation nP_{k} =\frac{n!}{(n-k)!}

10P_{4} =\frac{10!}{(10-4)!}=10.9.8.7.6= 5040

So, the probability that there are no repeated digits is:

P(A)=favorable cases/possible cases

P(A)=\frac{5040}{10000}=0.504 or 50.4%

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4 years ago
F(x) = x^3 + 3x^2<br><br> What would the graph look like?
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Answer: the answer is in the attachment.

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