If y varies jointly as x and w and y = 72 for x = 9 and w = 4, find the value of y for x = 18 and w = 1.5.

MATHEMATICS

Given:

If y varies jointly as x and w and y = 72 for x = 9 and w = 4, find the value of y for x = 18 and w = 1.5.

Solution:

y = 72 , x = 9 , w = 4

• y = kxw

• 72 = k(9)(4)

• 72 = k(36)

• k = 72/36

k = 2

Solve for the y.

x = 18 , w = 1.5

• y = 2xw

• y = 2(18)(1.5)

• y = 2(27)

y = 0.07Therefore, the answer is:

˚₊· ͟͟͞͞➳❥ [CarryOnLearning]

✎yoona18

2-1236-24249 – 3{6 – 10 + 6} = 9 – 6 = 3100 = 2  ´  2  ´  5  ´  558 = 2  ´  2937 is not the  product  of any two natural numbers (except trivially 37 = 1  ´  37) so it is called a  prime number.–6 = (-2)  ´  3   or   -6 = 3  ´  (-2)   or 
-6 = (-3)  ´  2   or   -6 = 2  ´  (-3)

Confidence interval for the population mean
z-test or normal distribution
solution:
given:  
n=20, x̄ =89, s=2.0, 99% 
formula:
x̄ -z_alpha/2(s/square root of n)< µ< x̄ +z_alpha/2(s/square root of n)
89-(2.575)(2.0/square root of 20)< µ< 89-(2.575)(2.0/square root of 20)
87.848< µ< 90.152

margin of error:
solution:
the critical value is 2.575.
the standard deviation is 2.0 (from the question), but as this is a sample, we need the standard error for the mean. 
the formula for the se of the mean is  standard deviation / √(sample size), so: 2.0/ √(20)=0.447.
me=2.575 * 0.447 = 1.151.

therefore, the margin of error is 1.151  with the confidence interval of 87.848< µ< 90.152.

Please provide details or choices.