Find the quotient. Simplify if possible. 7/9 divided by 5/6

Mathematics

2 answers:

Llana [10]3 years ago

6 0

1. (7/9)/(5/6)=14/15
2. 4/.67=6

Naddik [55]3 years ago

3 0

1. 7/9÷5/6
do reciprocal and now you are multiplying

7/9×6/5

7×6=42
9×5=45

now your fraction is 42/45
now divide 42 and 45 by 3

42÷3=14
45÷3=15

your simplified fraction is 14/15

2. 4/1÷2/3
do reciprocal again like the first one

4/1×3/2

4×3=12
1×2=2

your new fraction is 12/2, now divide 12 and 2 by 2

12÷2=6
2÷2=1

your simplified answer is 6

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Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

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s = 3.9

Confidence level = 95%

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$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$