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MatroZZZ [7]
2 years ago
3

Mrs. Carter needs 1/2 meter of cloth to make a pillowcase. How many pillowcases can she make with 5 meters of cloth?​

Mathematics
1 answer:
svp [43]2 years ago
4 0
She can make 10 pillow cases. 1/2 = p. So 5 / (1/2) = gives you the answer. Thus 10. Also can check by setting 5=10p. Then solving for p. P will equal 1/2.
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A high school student volunteers to present a report to the administration about the types of lunches students prefer. He survey
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<span>Convenience sampling.</span>
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2 years ago
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2 Rita bought three and forty-eight hundredths pounds of bananas at the store. How is this number written in expanded notation?
boyakko [2]

The cost of bananas = $ 3.48

This can be written as :                                                                                            

this number written in expanded notation as : (3 × 1) + (4 × 0.1) + (8 × 0.01)

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Hence, 1st option is correct.                                                  

5 0
2 years ago
there are two identical fridges for desserts. one of the fridge has 5/8 of its room left and the other fridge is only 1/8 full.
ELEN [110]

Answer:

1.5

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6 0
2 years ago
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33.32 divided by 9.8
kifflom [539]

Answer:

33.32 / 9.8 = 3.4;

Step-by-step explanation:


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2 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(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
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