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Artist 52 [7]
2 years ago
4

Maria lines up fabric squares in rows to make a quilt. She uses the same repeating pattern for all the rows. The rule is "Red, W

hite, Blue, White, Floral." If there are 125 quilt squares in the quilt, how many floral quilt squares will Maria need? Enter your answer in the box.
Mathematics
1 answer:
Katarina [22]2 years ago
6 0

Answer: 15

Step-by-step explanation:

the patern is

red

white

blue

white

floral.

So in each group of 5, we have one floral quilt square.

now, in 125 quilt squares, we want to find the number of groups of 5 that we can make with thos 125 quilt squares.

This is 125/5 = 15

So we have 15 groups of 5 quilt square.

And we know that in each of those groups, we have one floral quilt square.

So we have a total of 15*1 = 15 floral quilt squars.

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Function L is represented by the equation y 6x +4. Compare Functions B an determining which one has the greater rate of change a
Alenkasestr [34]

Answer:

-1

Step-by-step explanation:


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2 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)

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MODELING REAL LIFE In 60 seconds, a car in a parade travels 0.2 mile. The car traveled the last 0.05 mile in 12 seconds. How lon
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It is 4,000. :P
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