<span> The product of two perfect squares is a perfect square.
Proof of Existence:
Suppose a = 2^2 , b = 3^2 [ We have to show that the product of a and b is a perfect square.] then
c^2 = (a^2) (b^2)
= (2^2) (3^2)
= (4)9
= 36
and 36 is a perfect square of 6. This is to be shown and this completes the proof</span>
For the answer to the question above, I believe the answer is simply <u><em>8.
</em></u>
2 groups divided into four participants. So all in all people needed is 8.
I hope this helped you. Have a nice day!
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Answer:
y = 2.31x + 309.35
Step-by-step explanation:
Whenever you are checking for a best fitting equation you want to check if it has a constant slope. If it does then the relation is linear and super easy.
So, since the t values are increasing by the same amount you want to see if the y values are too. And they are, each population entry is increasing by 2.31, this is the slope.
Also, keep in mind you can caluclate slope with the equation 
here x is replaced by t though.
Now, since we know it's linear and we know the slope we can find the equation with the formula y - y1 = m(x - x1) where again, x is replaced by t and m is the slope 2.31
I am just going to use the first point. so x1 = t1 = 0
y - 309.35 = 2.31(x-0)
y = 2.31x + 309.35
Let me know if there was something you didn't understand.
Answer:
95% confidence interval for the proportion of students supporting the fee increase is [0.767, 0.815]. Option C
Step-by-step explanation:
The confidence interval for a proportion is given as [p +/- margin of error (E)]
p is sample proportion = 870/1,100 = 0.791
n is sample size = 1,100
confidence level (C) = 95% = 0.95
significance level = 1 - C = 1 - 0.95 = 0.05 = 5%
critical value (z) at 5% significance level is 1.96.
E = z × sqrt[p(1-p) ÷ n] = 1.96 × sqrt[0.791(1-0.791) ÷ 1,100] = 1.96 × 0.0123 = 0.024
Lower limit of proportion = p - E = 0.791 - 0.024 = 0.767
Upper limit of proportion = p + E = 0.791 + 0.024 = 0.815
95% confidence interval for the proportion of students supporting the fee increase is between a lower limit of 0.767 and an upper limit of 0.815.
Answer:
B) 11 green pepper plants
