When peanuts are packaged in their shells, most shells have two peanuts inside, but some have three and some only have one. A quality control inspector wants to determine if more than 10% of all shells contain just one peanut. If the inspector were to conduct a hypothesis test for p, the proportion of shells with just one peanut, what would be the appropriate pair of hypotheses for the inspector to use? H0: p = 0.1 and Ha: p ≠ 0.1 H0: p = 0.1 and Ha: p < 0.1 H0: p = 0.1 and Ha: p > 0.1 H0: p < 0.1 and Ha: p > 0.1 H0: p > 0.1 and Ha: p = 0.1

H0: p = 0.1 and Ha: p > 0.1

Step-by-step explanation:

At the null hypothesis, we test if the mean or proportion is equal to a certain value.

At the alternate hypothesis, we test if the mean or proportion is less than, more than, or different of the value tested at the null hypothesis.

A quality control inspector wants to determine if more than 10% of all shells contain just one peanut.

The value of 10% = 0.1 is tested, which means that the null hypothesis is

Test if it is more than 10% = 0.1, so the alternate hypothesis is:

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

76%

step-by-step explanation:

24% are dogs and the remaining is 76% because 100-24=76