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2 years ago

Find an angle between 0 and 2π that is coterminal with 10pi/3

Mathematics

1 answer:

azamat2 years ago

8 0

In this question we have to find an angle between 0 and 2π that is coterminal with 10pi/3 .

To find the coterminal angle, we need to subtract 2 pi from 10 pi/3. That is

$= \frac{10 \pi }{3} - 2 \pi$

Multiplying numerator and denominator of 2 pi by 3

$= \frac{10 \pi}{3} -\frac{6 \pi}{3} = \frac{4 \pi}{3}$

And that's the required coterminal angle between 0 and 2pi.

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Solve for c: c / 3 equals 30 / 5 show your work and apply the correct order of operation

Goshia [24]

Answer:

Unless there is more to the question than c = 2 because 30 /5 = 6 and if c is 3 than 3 = 6 so simplify if needed.

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2 years ago

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The greater of two consecutive EVEN numbers is 40 less than twice the smaller number. Find the larger number

Bas_tet [7]

Answer:

Step-by-step explanation:

z - some integer

2z - first given even integer (the smaller one)

2z+2 - even integer consecutive to 2z (the larger one)

2(2z) - twice the smaller number

2(2z) - 40 - the number 40 less than twice the smaller number

2(2z) - 40 = 2z + 2

4z - 40 = 2z + 2

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2z+2 = 42+2 = 44

check:

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3 years ago

What is thirty times twenty

wariber [46]

Answer:

600

Step-by-step explanation:

How to do it in your head (secret):

3*2 = 6

add 2 zeros after the 6 since there are 2 zeros in total in this equation 20 and 30 so:

30*20 = 600

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3 years ago

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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)

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2 years ago

Solve the x−4<2 or 4x−1>4 and write the solution in interval notation

jonny [76]

See photo. Hope it helps

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