Factor out 18.

The factors for 18 are as listed:

1, 2, 3, 6, 9, 18

Now factor out 30.

The factors for 30 are as listed:

1, 2, 3, 5, 6, 10, 15, 30

Now look at the factors that are both factors for 18 and 30.

So, the common factors are 2, 3, and 6.

5 0

3 years ago

The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding

Andru [333]

Answer:

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Step-by-step explanation:

Values (x) ∑(Xi-X)^2

----------------------------------

29.1 0.1702

28.5 1.0252

28.8 0.5077

29.4 0.0127

29.8 0.0827

30.1 0.3452

30.6 1.1827

----------------------------------------

236.1 3.4088

Mean = 236.1 / 8 = 29.51

$S_{x}=\sqrt{3.4088/(8-1)}=0.6978$

Statement of the null hypothesis:

H0: u ≥ 30 the mean wedding cost is not less than $30,000

H1: u < 30 the mean wedding cost is less than $30,000

Test Statistic:

$t=\frac{X-u}{S/\sqrt{n}}=\frac{29.51-30}{0.6978/\sqrt{8}}= \frac{-0.49}{0.2467}=-1.9861$

Test criteria:

SIgnificance level = 0.05

Degrees of freedom = df = n - 1 = 8 - 1 = 7

Reject null hypothesis (H0) if

$t$

Finding in the t distribution table α=0.05 with df=7, we have

$t_{0.05,7}=2.365$

$t>-t_{0.05,7}$ = -1.9861 > -2.365

Result: Fail to reject null hypothesis

Conclusion: Do no reject the null hypothesis

u ≥ 30 the mean wedding cost is not less than $30,000

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Hope this helps!

5 0

2 years ago