The size of the sample they should take to estimate p with a 2% margin of error and 90% confidence is n = 1691.
In statistics, the margin of error is just the degree of a significant error in the outcomes of random sample surveys.
The formula of margin error is, E = z√((p-vector)(1 - (p-vector)) ÷ n)
E = 2% = 0.02
Confidence level = 90%
Now, the proportion is not given so adopt nominal (p-vector) = 0.05
The critical value at CL of 90% is 1.645.
Thus, making n the subject,
n = z²(((p-vector) × (1 - (p-vector))) ÷ E²)
n = 1.645²((0.5 × 0.5) ÷ 0.02²)
n = 1691.266
n ≈ 1691
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Answer:
The answer to your questions are:
15.- 45 mg
16.- 25 kg
17.- 185 pounds
18.- 6 mg
Step-by-step explanation:
D = (a M) / (a + 12)
15.-
a = 4 years
M = 180
D = (4 x 180) / (4 + 12) = 720 / 16
D = 45 mg
16.- 1 pound --------------- 0.4535 kg
55 pounds ------------ x
x = 55 x 0.4535 / 1
x = 24.94 ≈25 kg
17.- 1 pound --------------- 0.4535 kg
x --------------- 84.09 kg
x = 84.09x 1 /0.4535
x = 185.42 pounds
18.- M = 18 mg what is 1/3 of the adult dose?
M = 18/3 = 6 mg
<span>The answer is B. C(n) = 0.75n - 0.25.
Let n be the number of pieces. The price for 1 piece is $0.75. The price C for n pieces without a coupon is n * $0.75: C(n) = 0.75n. The coupon value is $0.25. So, this value must be subtracted from the total price of n pieces. Since the coupon values in independent on the number of pieces, the price C for n pieces with the coupon will be: C(n) = 0.75n - 0.25. Therefore, the correct choice is B.</span>
The given above may be modeled by the arithmetic sequence with initial value (I) 2200 and common difference (d) of 70. The number of applicants every year can be written by the equation,
at = a1 + (n - 1) x d
From the given above, n is equal to 4. This corresponds to the term which is 3 years from now.
at = 2200 + (4 - 1) x 70 = 2410
Thus, the enrollment capacity would be 2410 students.
