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lesya692 [45]
2 years ago
6

I need the answer to this problem

Mathematics
1 answer:
Leya [2.2K]2 years ago
5 0

Answer:

116

Step-by-step explanation:

i answered most of your question in question comment section .....

hope you see before it gonna be deleted

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What is the y-/intercept of 5x+3y=15
liberstina [14]

Answer:

5

Step-by-step explanation:

y intercept: when x = 0

5(0) + 3y = 15

3y = 15, y = 5

The y intercept is 5

4 0
3 years ago
3x-2y=-1 and 3x-2y=9 solve using elimination
luda_lava [24]
You can put them vertically
then subtract them
but if you do it, it will be 0=-10 that doesn't make sense. 
So the answer is that there is no answer!
3 0
4 years ago
( -6, 8 ) is it a solution yes or no ? ​
Gennadij [26K]
Whats the original question you got from your teacher
8 0
3 years ago
A radioactive gas with a half-life of 2.4 days accidentally leaks into a room. If the radiation level is now 25% above a safe am
Gnom [1K]
Radioactive half-life is the time it takes for half an amount of radioactive material to decay into something else. In this case, it is assumed that the decay product is not radioactive or otherwise hazardous.

We must use the radioactive decay formula to determine at what time the radiation reaches a safe level.
A = Ao[e^(-0.693)(t)(t 1/2) where t 1/2 is the half-life, t is elapsed time, Ao is the original quantity, A is the future quantity.

We are given a half-life of 2.4 days , an Ao of 1.25 and an A of 1.00:

1.00 = (1.25)e^(-0.693)(2.4)t
1.00/1.25 = e^(-1.6632)t
0.8 = e^(-1.6632)t
t = 0.135 days = 3 hrs 15 min

This is the amount of time to a "safe level" using only radioactive decay, not venting or other means.
8 0
3 years ago
The marketing director of a large department store wants to estimate the average number of customers who enter the store every f
Dahasolnce [82]

Answer:

36.5674\leq x'\leq61.4326

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, \alpha is equal to 0.05 (it is calculated as: 1 - 0.95) and  t_{\alpha /2} is the t value with n-1 degrees of freedom that let a probability of \alpha/2 on the right tail.

So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and t_{\alpha /2} by 2.2621 we get:

49-2.2621\frac{17.38}{\sqrt{10} } \leq x'\leq 49+2.2621\frac{17.38}{\sqrt{10} }\\49-12.4326\leq x'\leq 49+12.4326\\36.5674\leq x'\leq61.4326

Finally, the interval values that she get is:

36.5674\leq x'\leq61.4326

8 0
3 years ago
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