For which sample size (n) and sample proportion (6) can a normal curve be used to approximate the sampling distribution?
A. n = 35; 6 = 0.9
B. n = 35; 6 = 0.8
C. n = 65; j = 0.9
D. n = 65; = 0.8

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

if the probability is 1/12, 75 people will receive a $25 gift card

step-by-step explanation:

step-by-step explanation:

a cube has 6 sides

the probability of getting 5 on the first cube is 1/6 since only one side of the cube has 5 on it.

there are 3 sides of the cube with even number on it [2,4,6]

therefore the probability of getting even number is 3/6

for both events to happen together, the probability would be

1/6 x 3/6 = 12/36

which simplifies to 1/12