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4 years ago

Please help me I really don’t get this so please help me

Mathematics

1 answer:

ludmilkaskok [199]4 years ago

3 0

Answer:

Step-by-step explanation:

As an example, let's divide 0.75 by 0.15. Multiply both numerator and denom. by 100, obtaining 75 / 15. Here we don't have to worry about decimal points any longer. Using a calculator or long division, you can show that 75 / 15 = 5, which is the same result if you "did" 0.75 / 0.15.

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Use the graph to find the x-intercept for the equation ý +2 Graph: Y + 2 = 2(x+3)

alex41 [277]

Answer:

(-2, 0)

Step-by-step explanation:

8 0

3 years ago

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One number is 4 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 538,

larisa [96]

Answer:

292

173

Step-by-step explanation:

x = first number

4x = second number

100 + x = third number

x + 4x + 100 + x = 538

6x + 100 = 538

6x + 100 - 100 = 538 - 100

6x = 438

6x/6 = 438/6

x = 73

Check:

73 + 4(73) + 100 + 73 = 538

73 + 292 + 173 = 538

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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