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4 years ago

PLZ HELP GEOMETRY BELOW

Mathematics

1 answer:

Nimfa-mama [501]4 years ago

4 0

The answer is C because AO and BO are equal. And with AO being three times the length of CO, BO should be the length of DO

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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

Solve the equation. \dfrac32 + b = \dfrac74 2 3 +b= 4 7 start fraction, 3, divided by, 2, end fraction, plus, b, equals, sta

Xelga [282]

Given:

The equation is

$\dfrac{3}{2}+b=\dfrac{7}{4}$

To find:

The value of b.

Solution:

We have,

$\dfrac{3}{2}+b=\dfrac{7}{4}$

Subtracting both sides by $\dfrac{3}{2}$ , we get

$b=\dfrac{7}{4}-\dfrac{3}{2}$

$b=\dfrac{7-2(3)}{4}$

$b=\dfrac{7-6}{4}$

$b=\dfrac{1}{4}$

Therefore, the value of b is $\dfrac{1}{4}$ .

3 0

3 years ago

Read 2 more answers

2. How tall is the flagpole? Use a proportion.

Reptile [31]

7.5 ft

I dont think so.. I think thats the answer

8 0

3 years ago

Write in intercept form

Alex787 [66]

-x² - x + 6

First thing I would do is distribute out that negative. Your new problem is:

- (x² + x - 6)

Now, factor the inside. We need to find products of - 6 that when added together have a sum of positive 1. They are 3 and - 2.

- (x² + 3x - 2x - 6)

Group together and find your factors. I'm going to short cut and go right to them for the sake of space and time.

- (x + 3)(x - 2)

Your answer is Option B

7 0

3 years ago

Help with this pleaseee

BartSMP [9]

Answer:

29 units²

Step-by-step explanation:

You could divide this into two triangles and two trapezoids to find the area. But since the majority of these are either whole or half squares, I‘d count those. Then you’d just have one triangle in the top right corner.

Whole squares: 21

Half Squares: 10 (so 4 square units)

Triangle: 1/2(b)(h)=1/2(5)(2)=5

TOTAL = 21+4+5= 29 units²

7 0

3 years ago