Answer:
Hypothesis rejected
Step-by-step explanation:
Lets use the t-test since the variance of the population is now known. We need to test the hypothesis that H_0: \mu \leq 79 \text{ vs } H_1: \mu > 79 . This is performed in R as follows:
t.test(tt$Test.2.Score,mu=79,alternative="greater")
One Sample t-test
data: tt$Test.2.Score
t = 2.9238, df = 69, p-value = 0.002337
alternative hypothesis: true mean is greater than 79
95 percent confidence interval:
81.26555 Inf
sample estimates:
mean of x
84.27143
Thus, we reject the null hypothesis and conclude that \mu > 79.
The correct answer is 70.08 all you have to do is multiply them together.
Given data :
a₃ = 9/16
aₓ = -3/4 · aₓ₋₁
Where x is the number of terms ('x' is also written as 'n')
To find the 7th term (a₇):
We know that aₓ = -3/4 · aₓ₋₁
So,
a₃ = -3/4 · a₃₋₁
a₃ = -3/4 · a₂
9/16 = -3/4 · a₂
a₂ = 9/16 × -4/3
a₂ = -36/48
a₂ = -3/4
Again,
aₓ = -3/4 · aₓ₋₁
a₄ = -3/4 · a₄₋₁
a₄ = -3/4 · a₃
a₄ = -3/4 · 9/16
a₄ = -27/64
a₄ = -27/64
For a₅,
aₓ = -3/4 · aₓ₋₁
a₅ = -3/4 · a₅₋₁
a₅ = -3/4 · a₄
a₅ = -3/4 × -27/64
a₅ = 81/256
For a₆,
aₓ = -3/4 · aₓ₋₁
a₆ = -3/4 · a₆₋₁
a₆ = -3/4 · a₅
a₆ = -3/4 × 81/256
a₆ = -243/1024
For a₇,
aₓ = -3/4 · aₓ₋₁
a₇ = -3/4 · a₇₋₁
a₇ = -3/4 · a₆
a₇ = -3/4 × -243/1024
a₇ = 729/4096
Answer:
Number of weeks in a year = 52 weeks
If in a normal year when each package of flea treatment lasts for 4 weeks, then in a year there Jim's dog will have to be treated for
Where as, when the fleas are bad in a year, the treatment lasts for only 3 weeks.
Then in a year Jim's dog would get
So Jim's dog will get 17- 13 =4 treatments more.
4 treatments that are made in 3 weeks each will be 4×3 =12 weeks more treatment
The moon does. The moon is waaaay bigger than any of these combined.