The answer is the 3rd one
Amount of money collected in the 2-day fundraiser is represented by the following equations:
- 12x + 4y = 152 .................. (i)
- 32x + 12y = 420 .................. (ii)
where x is the cost of a chicken lunch and y is the cost of a vegetarian lunch
Multiplying (i) with 3 and subtracting it from (ii)
36x + 12y - (32x + 12y) = 456 - 420
⇒ 4x = 36
⇒ x = 9
Put value of x in (i)
⇒ 12 × 9 + 4y = 152
⇒ 108 + 4y = 152
⇒ 4y = 44
⇒ y = 11
Hence, cost of the vegetarian meal is 11.
Answer:
95% z-confidence interval for the proportion of all children enrolled in kindergarten who attended preschool is between a lower limit of 0.528 and an upper limit of 0.772.
Step-by-step explanation:
Confidence interval = p + or - zsqrt[p(1-p) ÷ n]
p is sample proportion = 39/60 = 0.65
n is the number of children sampled = 60
Confidence level (C) = 95% = 0.95
Significance level = 1 - C = 1 - 0.95 = 0.05
Divide significance level by 2 to obtain critical value (z)
0.05/2 = 0.025 = 2.5%
z at 2.5% significance level = 1.96
zsqrt[p(1-p) ÷ n] = 1.96sqrt[0.65(1-0.65) ÷ 60] = 1.96sqrt[0.2275 ÷ 60] = 1.96sqrt(3.792×10^-3) = 1.96×0.062 = 0.122
Lower limit = p - 0.122 = 0.65 - 0.122 = 0.528
Upper limit = p + 0.122 = 0.65 + 0.122 = 0.772
95% confidence interval is between 0.528 and 0.772
Complete question :
A zoologist is interested in whether climate affects how high kangaroos can jump. In a random sample of gray kangaroos from an Australian forest, he found the mean jump height was x¯=286 inches with a margin of error of 15 inches.
Construct a confidence interval for the mean jump height of gray kangaroos.
Answer:
(271, 301)
Step-by-step explanation:
Given that:
Mean (x¯) = 286
Margin of Error = 15 inches
The confidence interval :
(Mean - Margin of Error), (mean + margin of error)
(286 - 15), (286 + 15)
(271, 301)
Hence confidence interval equals : (271, 301)
