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Amiraneli [1.4K]
3 years ago
8

How do I do this I don't get it

Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
8 0
Just do what you would do with any other number (real numbers such as 1,2 ETC...)
multiplication, subtraction, addition, division, and exponents.
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-7/10+7/15+1/-20+-9/10+11/15+11/-20 <br> what is the answer please help me .....!!!!
Nat2105 [25]

Answer:

-1

Step-by-step explanation:

-7/10+7/15+1/-20+-9/10+11/15+11/-20

= -7/10 + -9/10 + 7/15 + 11/15 + 1/-20 + 11/-20

= -16/10 + 18/15 + 12/-20

= -8/5 + 6/5 + -3/5

= (-8+6-3)/5

= -5/5

= -1

7 0
3 years ago
A number has the same digit in its hundreds place and its hundredths place. How many times greater is the value of the digit in
Free_Kalibri [48]

Answer:

A number has the same digit in its hundreds place and its hundredths place. Let the digit be '1'. Hence, the the value of the digit in the hundreds place is 10,000 times greater than the value of the digit in the hundredths place

Step-by-step explanation:

5 0
3 years ago
What he the value of n?<br> 3n+16=7n
Vlad [161]

Answer:

The answer is 4

Step-by-step explanation:

3n+16=7n

collect like terms

-4n=-16

divide by -4

n=4

5 0
4 years ago
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dimaraw [331]
Hello to Mr Fowler's science class from brainly!!!
3 0
3 years ago
The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 4.5 per week. Find the probabili
Mariana [72]

Answer:

a) 0.01111

b) 0.4679

c) 0.33747

Step-by-step explanation:

We are given the following in the question:

The number of accidents per week can be treated as a Poisson distribution.

Mean number of accidents per week = 4.5

\lambda= 4.5

Formula:

P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}

a) No accidents occur in one week.

P(x =0)\\\\= \displaystyle\frac{4.5^0 e^{-4.5}}{0!}= 0.01111

b) 5 or more accidents occur in a week.

P( x \geq 5) = 1-\displaystyle \sum P(x

c) One accident occurs today.

The mean number of accidents per day is given by

\lambda = \dfrac{4.5}{7} = 0.64

P(x =1)\\\\= \displaystyle\frac{0.64^1 e^{-0.64}}{1!}= 0.33747

5 0
3 years ago
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