We take the equation <span>d = -16t^2+12t</span> and subtract d from both sides to get
0<span> = -16t^2+12t - d
We apply the quadratic formula to solve for t. With a = -16, b = 12, c = -d, we have
t = [ -(12) </span><span>± √( 12^2 - 4(-16)(-d) ) ] / [2 * -16]</span>
= [- 12 ± √(144-64d) ] / (-32)
= [- 12 ± √16(9-4d)] / (-32)
= [- 12 ± 4√(9-4d)] / (-32)
= 3/8 ±√(9-4d) / 8
The answer to your question is t = 3/8 ±√(9-4d) / 8
Answer:
20-8 = 12 , he waste $12 if he got 4 cans 12 divide by 4 equal 3 so the one can cost $3
Step-by-step explanation:
Answer:
24, 48, 72, 96, 120
Step-by-step explanation:
I can keep going on but i learned to just put at least 5 and if the person needs more then just give them ten more
Answer:
The value is 
Step-by-step explanation:
From the question we are told that
The population proportion is 
The sample size is n = 563
Generally the population mean of the sampling distribution is mathematically represented as

Generally the standard deviation of the sampling distribution is mathematically evaluated as

=>
=>
Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

Here
is the sample proportion of persons with a college degree.
So
![P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )](https://tex.z-dn.net/?f=P%28%20-%20%280.05%20-%200.52%20%29%20%3C%20%20%5C%5E%20p%20%3C%20%20%280.05%20%2B%200.52%20%29%29%20%3D%20P%28%5Cfrac%7B%5B%5B0.05%20-0.52%5D%5D-%200.52%7D%7B0.02106%7D%20%3C%20%5Cfrac%7B%5B%5C%5Ep%20-%20p%5D%20-%20p%7D%7B%5Csigma%20%7D%20%20%3C%20%5Cfrac%7B%5B%5B0.05%20-0.52%5D%5D%20%2B%200.52%7D%7B0.02106%7D%20%29)
Here
![\frac{[\^p - p] - p}{\sigma } = Z (The\ standardized \ value \ of\ (\^ p - p))](https://tex.z-dn.net/?f=%5Cfrac%7B%5B%5C%5Ep%20-%20p%5D%20-%20p%7D%7B%5Csigma%20%7D%20%20%3D%20Z%20%28The%5C%20standardized%20%5C%20%20value%20%5C%20%20of%5C%20%20%28%5C%5E%20p%20-%20p%29%29)
=> ![P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 } < Z < \frac{-0.47 + 0.52}{0.02106 }]](https://tex.z-dn.net/?f=P%28%20-%20%280.05%20-%200.52%20%29%20%3C%20%20%5C%5E%20p%20%3C%20%20%280.05%20%2B%200.52%20%29%29%20%3D%20P%5B%5Cfrac%7B-0.47%20-%200.52%7D%7B0.02106%20%7D%20%20%3C%20%20Z%20%20%3C%20%5Cfrac%7B-0.47%20%2B%200.52%7D%7B0.02106%20%7D%5D)
=> ![P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[ -2.37 < Z < 2.37 ]](https://tex.z-dn.net/?f=P%28%20-%20%280.05%20-%200.52%20%29%20%3C%20%20%5C%5E%20p%20%3C%20%20%280.05%20%2B%200.52%20%29%29%20%3D%20P%5B%20-2.37%20%3C%20%20Z%20%20%3C%202.37%20%5D)
=> 
From the z-table the probability of (Z < 2.37 ) and (Z < -2.37 ) is

and

So
=>
=>
=> 
<h2>~<u>Solution</u> :-</h2>
Here, it is given that the bag contains 25 paise coins and 50 paise coins in which, 25 paise coins are 6 times than that of 50 paise coins. Also, the total money in the bag is Rs. 6.
- Hence, we can see that, here, we have been given the linear equation be;
Let the number of coins of 50 paise will be $ x $ and the number of coins of 25 paise will be $ 6x $ as given. . .
Hence,




- Hence, the number of 50 paise coins will be <u>2</u>. And, 6 times of two be;

- Hence, the number of 25 paise coins will be <u>12</u>.