Solve for y in 3(2y-4)=8y+9-9y

HELP MEEEEE!!!!!!!!!!!

Nina [5.8K]

Answer:

B. ii and iii

Sorry if im wrong

Step-by-step explanation:

6 0

2 years ago

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

2 years ago

what is the distance around a triangle that has sides measuring 2 1/8 feet, 3 1/2 feet, and 2 1/2 feet?

Vlad1618 [11]

7 and 1/8. all you need to do is add these together by getting a common denominator of eight

5 0

2 years ago

Read 3 more answers

Write the equation 6561 1/4= 9 in logarithmic form.

Vlad1618 [11]

Answer:

b. log6561(9) = 1/4

Step-by-step explanation:

The relationship between the exponential form and the log form is ...

$b^n=x\\\log_b{(x)}=n$

The exponent is the logarithm. The base is the value the exponent is being applied to.

Here, you have ...

$6561^{\frac{1}{4}}=9$

Comparing this to the form above, we see ...

b = 6561, n = 1/4, x = 9

so the log form of the equation is ...

$\boxed{\log_{6561}{(9)}=\frac{1}{4}}$

This matches choice B.

8 0

2 years ago

A batch of 445 containers for frozen orange juice contains 3 that are defective. Two are selected, at random, without replacemen

Elan Coil [88]

Answer:

When Two containers are selected

(a) Probability that the second one selected is defective given that the first one was defective = 0.00450

(b) Probability that both are defective = 0.0112461

(c) Probability that both are acceptable = 0.986

2. When Three containers are selected

(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451

(c) Probability that all three are defective = 6.855 x $10^{-8}$ .

Step-by-step explanation:

We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.

Total containers = 445

Defective ones = 3

Non - Defective ones = 442 { Acceptable ones}

Two containers are selected, at random, without replacement from the batch.

(a) Probability that the second one selected is defective given that the first one was defective is given by;

Since we had selected one defective so for selecting second the available

containers are 444 and available defective ones are 2 because once

chosen they are not replaced.

Hence, Probability that the second one selected is defective given that the first one was defective = $\frac{2}{444}$ = 0.00450

(b) Probability that both are defective = P(first being defective) +

P(Second being defective)

= $\frac{3}{445} + \frac{2}{444}$ = 0.0112461

(c) Probability that both are acceptable = P(First acceptable) + P(Second acceptable)

Since, total number of acceptable containers are 442 and total containers are 445.

So, Required Probability = $\frac{442}{445}*\frac{441}{444}$ = 0.986

Three containers are selected, at random, without replacement from the batch.

(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;

Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .

Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = $\frac{1}{443}$ = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;

Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.

Hence, Required probability = $\frac{2}{443}$ = 0.00451 .

(c) Probability that all three are defective = P(First being defective) +

P(Second being defective) + P(Third being defective)

= $\frac{3}{445}* \frac{2}{444} * \frac{1}{443}$ = 6.855 x $10^{-8}$ .

5 0

2 years ago