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Mrrafil [7]
4 years ago
7

Which of the following eaquals 6000000 in scintific notation

Mathematics
2 answers:
Gemiola [76]4 years ago
8 0
6×10to the sixth power
sukhopar [10]4 years ago
6 0
Count the number of 0's after the 6  ( = 6)

answer is 6 * 10^6
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Simplify the expression <br><br> c+2+c+c+4
professor190 [17]

Answer:

2+4+ccc

Step-by-step explanation:

7 0
2 years ago
There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=
yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = \frac{12P4}{12^{4} }

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = \frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }

P(X=2) = 0.401

P(X=3) = \frac{4C3 * 12P2}{12^{4} }

P(x=3) = 0.025

P(X=4) = \frac{12}{12^{4} }

P(X=4) = 0.001

Mean, \mu = \sum xP(x)

\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879

Standard deviation, SD = \sqrt{\sum x^{2} P(x) - \mu^{2}}  \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2}  * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}

SD = 1.033

4 0
4 years ago
A store sells
user100 [1]
C
Hope this helps!!!!!
7 0
3 years ago
Y=3/2(-1)-1<br> Linear Equation
levacccp [35]
X = -5/2
X is a single horizontal line across the graph, because x has a set value

5 0
4 years ago
PLEASE HELP QUICK I CANT FAIL THIS TEST
Finger [1]

Answer:

24+24x I believe is the answer. Good luck on your test :)

5 0
4 years ago
Read 2 more answers
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