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PilotLPTM [1.2K]
2 years ago
6

Which decimal is equivalent to forty and seventy six thousands ? a.40.076 b 40.76 C 47.006 D 47.06​

Mathematics
1 answer:
krek1111 [17]2 years ago
7 0

Answer:

A. 40.076

Step-by-step explanation:

Because B is forty and seventy six hundredths. C is forty-seven and six thousandths. D is forty-seven and six hundredths.

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16 is 20% of what number?
ehidna [41]

Answer:

80

Step-by-step explanation:

80 * 20% = 16

then 16 is 20% of 80

4 0
3 years ago
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Which statement is true about the graphs of the two lines y = –6 and x = 1/6<br>?
ivanzaharov [21]
The lines are perpendicular to each other because the graph of y = –6 is a horizontal line with a slope that is undefined, and the graph of x = 1/6 is a vertical line with a slope of 0.
The lines are perpendicular to each other because the graph of y = –6 is a vertical line with a slope that is undefined, and the graph of x =1/6 is a horizontal line with a slope of 0.
The lines are perpendicular to each other because the graph of y = –6 is a vertical line with a slope of 0, and the graph of x =1/6 is a horizontal line with a slope that is undefined.
<span> The lines are perpendicular to each other because the graph of y = –6 is a horizontal line with a slope of 0, and the graph of x =1/6 is a vertical line with a slope that is undefined. </span>
4 0
3 years ago
Aaron scored 452.65 marks out of 600 in the final examination. How many marks did he lose?
lana [24]

Answer:

He lost 147.35 marks

Step-by-step explanation:

Take the total marks and subtract the marks he got to find the marks he lost

600-452.65= 147.35

7 0
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(T0 (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
If right answer good review
Nookie1986 [14]

Answer:

Bottom right graph is the correct answer

3 0
2 years ago
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