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tamaranim1 [39]
3 years ago
13

Sales fall from 300 per week to 270 per week what’s the percentage change

Mathematics
1 answer:
weeeeeb [17]3 years ago
8 0
Hi there! 

Your question: 

Sales fall from 300 per week to 270 per week what's the percentage change?

My answer:

The formula for calculating percent change is as follows: 

[(y2 - y1)/y1]*100=your percent change 

y2= first value 
y1=second value 

Plug the numbers in: 

[(300-270)/300]*100

[30/300]*100

0.1*100

10 

Therefore, the percent change is 10%

Hope this helps! Let me know if it's incorrect so I can fix it:)
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Answer:

9/12 or 75%.

Step-by-step explanation:

Let's first add the choices all up.

5+4+3= 12.

We have 12 movie choices in total.

Now, we know that we have 3 drama movie choices...so the chance we will choose drama is 3/12, but we are not finding that.  We want the chance she will NOT choose drama, so that would be 9/12.

*Since the chance she will choose drama is 3/12, the chance she will not choose is 9/12. 9+3 is 12, the total amout.*

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-7X&gt;10<br>solve for x answer musr be simplifyed<br>​
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Answer:

x < -10/7

Step-by-step explanation:

Divide both sides by -7.  Because this divisor is negative, we must reverse the direction of the inequality sign, obtaining:

-7x   <   10

-----   <  -----

 -7         -7

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One less than five times a number is fewer than twenty-four
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1 - 5x < 24

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What is the best next step in the construction of the perpendicular bisector <br> AB?
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A rare form of malignant tumor occurs in 11 children in a​ million, so its probability is 0.000011. Four cases of this tumor occ
Shkiper50 [21]

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

\lambda=r*t=0.000011*13,280=0.14608

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990

c) The probability of more than one case is:

P(k>1)=1-P(k\leq 1)=1-0.990=0.010

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

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3 years ago
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