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Pavel [41]
3 years ago
5

Which expression has a value equal to 28 divided 4

Mathematics
2 answers:
Morgarella [4.7K]3 years ago
8 0

Answer: Check explanation, could be anything.

Step-by-step explanation: This could honestly be anything but heres a few examples.

28/4

7

56/8

14/2

63/9

70/7

√49

∛343

krok68 [10]3 years ago
3 0

Answer:

Step-by-step explanation:

28 \ 4 = 32

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What is the completely factored form of this polynomial?<br> 81x4 − 16y4
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Answer:

( 9 x^ 2 + 4 y ^2 ) ( 3 x + 2 y ) ( 3 x − 2 y )

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On a coordinate plane, DEF has vertices D(1,2), E(7,2)and F(1,9). What is the distance between point E and F?
notka56 [123]

Step-by-step explanation:

Using distance formula,

Let the distance between point E and F = ef,

ef  =  \sqrt{{(7 - 1)}^{2}  +  {(2 - 9)}^{2} }  \\ ef =  \sqrt{85}  \\  = 9.219544457

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) Which list orders the fractions from least to greatest? A.) ½, ¾, ¼, ⅜ B.) ¾, ½, ⅜, ¼ C.) ¼, ⅜, ½, ¾ D.) ⅜, ½, ¾,
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3 years ago
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Dima020 [189]
The answer is 3 3/8 miles
3 0
3 years ago
Read 2 more answers
In 2011, New York City has a total of 11,232 motor vehicle accidents that occurred on Monday through Friday between the hours of
cestrela7 [59]

Answer:

a. \ P(X=0)=5.574\times10^{-7}

b. \ \ P(X\leq 1)=8.584\times10^{-6}

c.\ \ \ P(X\geq 4)=0.9997

Step-by-step explanation:

a. #We notice this is a Poisson probability function expressed as:

f(x)=\frac{\mu^xe^{-\mu}}{x!}

x-number of occurrences in a given interval.

\mu-mean occurrences of the event

-The mean is calculated as:

\mu=\frac{14.4}{4}=3.6

#the probability of no accidents in a 15-minute period is :

P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X=0)=\frac{14.4^0e^{-14.4}}{0!}\\\\=5.574\times10^{-7}

Hence, the probability of no accident in a 15-min period is =5.574\times10^{-7}

b. The the probability of at least one accident in a 15-minute period. is calculated as:

P(x)=\frac{\mu^xe^{-\mu}}{x!}\\\\P(X\leq 1)=\frac{14.4^0e^{-14.4}}{0!}+\frac{14.4^1e^{-14.4}}{1!}\\\\=5.574\times10^{-7}+8.026\times 10^{-6}\\\\=8.584\times 10^{-6}

Hence,  the probability of at least one accident in a 15-minute period is 8.584\times10^{-6}

c. The probability of four or more accidents in a 15-minute period is calculated as:

P(X\geq 4)=1-P(X\leq 3)=1-[P(X=0)+P(X1)+P(X=2)+P(X=3)]\\\\=1-[5.574\times10^{-7}+a. \ 8.026\times10^{-6}+\frac{14.4^2e^{-14.4}}{2!}+\frac{14.4^3e^{-14.4}}{3!}]\\\\=1-[8.584\times 10^{-6}+5.779\times10^{-5}+2.774\times10^{-4}]\\\\=0.9997

Hence,the probability of four or more accidents in a 15-minute period. is  0.9997

7 0
3 years ago
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