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spin [16.1K]
3 years ago
9

The total of 36 78 198 475 and 620

Mathematics
2 answers:
Alina [70]3 years ago
8 0
The total of,
  36
  78
 198
 475
+620
-------
=1407

Hope this helps!=)
levacccp [35]3 years ago
3 0
1407 is the total of 36, 78, 198, 475 and 620
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Answer:

On a coordinate plane, a graph titled Total Pages Jacqueline Has Read has days on the x-axis and Pages on the y-axis. A line goes through points (1, 25) and (2, 50).

Step-by-step explanation:

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Please solve 25/5(2+3)=
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10x-5 9x+2 is my equation and I don’t know how to solve it I really need help,someone please help me solve it.
Ostrovityanka [42]

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Answer:

  x = 7

Step-by-step explanation:

You solve a linear equation by putting the variable on one side of the equal sign and a constant on the other side. Here, variables and constants are on both sides of the equal sign, so you need to separate them.

The basic idea is that you add the opposite of any term you don't want. Whenever you perform any operation (like "add"), <em>you must do it to both sides of the equation</em>.

We observe that x-terms have coefficients of 10 and 9. We choose to add the opposite of 9x to both sides:

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  x -5 = 2 . . . . simplify

Now, we still have -5 on the left, where we don't want it. So, we add its opposite (+5) to both sides:

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The solution is x = 7.

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<em>Additional comment</em>

If we were to end up with an x-coefficient other than 1, we would divide both sides of the equation by that coefficient. This will leave the x-term with a coefficient of 1.

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2 years ago
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1 year ago
According to a study done by Nick Wilson of Otago University Wellington, the probability a randomly selected individual will not
Lorico [155]

Answer and Step-by-step explanation:

From the question statement we get know that it is Binomial distribution because there are only two possible outcomes so we need to use Binomial Probability Distribution for this question.

Formula for the Binomial Probability Distribution:

P(X)=   p^x q^(n-x)

Where,

  • C_x^n=n!/(n-x)!x!   (i.e. combination)
  • x= total number of successes
  • p=probability of success (p=1-q)  
  • q=probability of failure (q=1-p)
  • n=number of trials
  • P(X)= probability of total number of successes

Answer and explanation for each part of the question are as follow:

a.What is the probability that among 10 randomly observed individuals exactly 4 do not cover their mouth when sneezing?

Solution:

Given that

n=10  

p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)

q=1-0.267=0.733  

x=4 (number of successes i.e. individuals not covering their mouths)

C_x^n=n!/(n-x)!x!=10!/(10-4)!4!=210

P(X)=C_x^n   p^x q^(n-x)=210×〖(0.267)〗^4×〖0.733〗^(10-4)

P(X)=210×0.00508×0.155  

P(X)=0.165465  

b. What is the probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth when sneezing?

Solution:

Given that

n=10  

p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)

q=1-0.267=0.733  

x=3 (number of successes i.e. individuals not covering their mouths)

C_x^n=n!/(n-x)!x!=10!/(10-3)!3!=120

P(X)=C_x^n   p^x q^(n-x)=120×(0.267)^3×〖0.733〗^(10-3)

P(X)=120×0.01903×0.1136  

P(X)=0.25962  

c. Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? why?

Solution:

Given that

n=18  

p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)

q=1-0.267=0.733  

x=9 (x is the number of successes “number of individuals not covering their mouths when sneezing”, if less than half cover their mouth then more than half will not cover), so let x=9

C_x^n=n!/(n-x)!x!=18!/(18-9)!9!=48620

P(X)=48620×(0.267)^9×〖0.733〗^(18-9)  

P(X)=48620×0.00000689×0.0610  

P(X)=0.020  

Yes, I am surprised that probability of less than 9 individuals covering their mouth when sneezing is 0.020. Which is extremely is small.

3 0
3 years ago
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