Answer:
865
Step-by-step explanation:
We have that in 95% confidence level the value of z has a value of 1.96. This can be confirmed in the attached image of the normal distribution.
Now we have the following formula:
n = [z / E] ^ 2 * (p * q)
where n is the sample size, which is what we want to calculate, "E" is the error that is 2% or 0.02. "p" is the probability they give us, 5 out of 50, is the same as 1 out of 10, that is 0.1. "q" is the complement of p, that is, 1 - 0.1 = 0.9, that is, the value of q is 0.9.
Replacing these values we are left with:
n = [1.96 / 0.02] ^ 2 * [(0.1) * (0.9)]
n = 864.36
865 by rounding to the largest number.
Answer:
<em>No</em><em>.</em><em> </em><em>of</em><em> </em><em>sweets</em><em> </em><em>=</em><em> </em><em>123</em><em> </em>
<em>No</em><em>.</em><em> </em><em>of</em><em> </em><em>children</em><em> </em><em>=</em><em> </em><em>7</em><em> </em>
<em>then</em>
<em>sweets</em><em> </em><em>per</em><em> </em><em>children</em><em> </em><em>=</em><em>123</em><em>/</em><em>7</em><em> </em><em>=</em><em>17</em><em>.</em><em> </em><em>57</em>
(-5,2) concave up
Hope this helps :)
4 > 6*t
Number t refers to an unknown. In this case, t can only be 0, because 6*0 = 0, and 4 > 0 or four is greater than 0.