Does anyone know answer to this ?

. HELP DUE IN A FEW MIN ASAP im stupidssssss

adelina 88 [10]

Answer:

B

Step-by-step explanation:

80*17=1360/choice B

5 0

3 years ago

How do you solve what is 30% of $240

DerKrebs [107]

The Answer would be x=72. Why? Because $\frac{x}{240} = \frac{30}{100}. Fraction \frac{30}{100}$ is your %. You Cross multiply 240 * 30 = 100x
240 time 30 = 7200. you want x to be alone so you do 7200 divided by 100 equaling 72. So you get x=72.

6 0

3 years ago

In a recent golf match, Tiger’s score was 4 less than Phil’s score. Their combined scores totaled 140. Let p represent Phil's sc

Svetach [21]

The answer would be D .

3 0

3 years ago

Sasha had $1800 and share it in the ratio 2:3:4 between her 3 cousin what was the largest share

PilotLPTM [1.2K]

Answer:

$800

Step-by-step explanation:

$total \: units \: = \\ 2 + 3 + 4 = 9 \: units \\ 9 \: units \: = 1800 \\ 1 \: unit \: = 1800 \div 9 \: = 200 \\ largest \: share \: = 4 \: units \\ 4 \: units \: = \: 200 \times 4 = 800$

5 0

3 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

3 years ago