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dem82 [27]
3 years ago
13

Which expresion represents the perimeter of the rectangle above (3y-5), 2y

Mathematics
2 answers:
Ludmilka [50]3 years ago
6 0

Answer: The expression represents the perimeter is 10y-10.

Step-by-step explanation:

Let the length of rectangle be 3y-5

Let the breadth of rectangle be 2y

As we know the formula for perimeter,

So,

\text{Perimeter of rectangle }\\\\=2(L+B)\\\\=2(3y-5+2y)\\\\=2(5y-5)\\\\=10y-10

This is the required expression which represents the perimeter of the rectangle.

Sonbull [250]3 years ago
3 0
(3y-5 + 2y)*2 = (5y-5)*2 = 10y-10
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Sometimes it will, and sometimes it won't.

I reason as follows:

(21 + x)  +  (30 + 2x)  =  51 + 3x      has an 'x' term

(42 + x)  +  (30 - x)  =  72                  has no 'x' term.

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Explain why receiving a 15% discount is the same as finding 85% percent of the original cost of an item. Then, explain how the m
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Answer:

15% discount means you're paying 85% of the marked price because total is 100% and 100-15 = 85%

Let X be the naked price

Paid = marked price - discount

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3 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.

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(\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}}

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-0.03\pm(2.0167)(0.012459)(0.3), i.e.,

(-0.0225, -0.0375)

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