The sugar content of the syrup is canned peaches is normally distributed. assumethe can is designed to have standard deviation 5 milligrams. a random sample ofn= 10 cans is studied. what is the sampling distribution of the sample variance? the data yields a sample standard deviation of 4.8 milligrams. what is the chanceof observing the sample standard deviation greater than 4.8 milligrams?

0.50477

Step-by-step explanation:

Given : The sugar content of the syrup is canned peaches is normally distributed.

We assume the can is designed to have standard deviation milligrams.

The sampling distribution of the sample variance is chi-square distribution.

Also,The data yields a sample standard deviation of milligrams.

Sample size : n= 10

Test statistic for chi-square :

i.e.

Now, P-value =  [By using the chi-square distribution table for p-values.]

Hence, the chance of observing the sample standard deviation greater than 4.8 milligrams = 0.50477

b.-4 units

all you   do is count how many times you move

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x