All categories

4 years ago

Record Subtraction with Renaming

Mathematics

1 answer:

kotykmax [81]4 years ago

3 0

2. 2 2/5
3. 2 1/2
4. 1/2
5. 4 3/5
6. 3 1/3
7. 2 4/5
8. 2
9. 4 1/3
10. 4 2/3
11. 2/5
12. 8 2/3
13. 1 1/2
14. 2 3/4
15. 1 5/12
16. 5 2/5

You might be interested in

Please help me on this question.

scoundrel [369]

Answer:

$f(1)=-48(-\frac{1}{4} )$

$= 12$

$f(n)=f(n-1)\:. (-\frac{1}{4} )$

<u />

<u>OAmalOHopeO</u>

4 0

3 years ago

Read 2 more answers

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago

A tennis tournament has 2n contestants. We want to pair them up for the first round of singles matches. Show that the number of

ddd [48]

There are

$\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}$

ways of pairing up any 2 members from the pool of $2n$ contestants. Note that

$(2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)$

so that

$\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}$

4 0

2 years ago

Please help I need it ASAP

Varvara68 [4.7K]

Answer:

do 1+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=61+1=6

Step-by-step explanation:

5 0

3 years ago

Whats the inequalite of c-10>10

storchak [24]

Answer: c > 20

Step-by-step explanation:

add ten to both sides

4 0

3 years ago

Read 2 more answers