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e-lub [12.9K]
3 years ago
5

A new miniature golf and arcade opened up in town. For convenient ordering, a play package is available to purchase. It includes

two rounds of golf and 20 arcade tokens, plus $3.00 off the regular price. Stacey and her six friends are purchasing this package. Let g represent the cost of a round of golf, and let t represent the cost of a token. Write two different expressions that represent the total amount this group spent.
Mathematics
1 answer:
ruslelena [56]3 years ago
7 0

Answer:

a. 7(2g + 20t –3)

b. 14g + 140t – 21

Step-by-step explanation:

a. The first expression 7(2g + 20t  – 3)

Stacey and her six friends implies that we have 7 members in the group.

Therefore, the expression above implies that each member of the group has to pay for two rounds of golf and 20 tokens will be discounted $3.00.

The expression (2g + 20t –3) represents the quantity of cost of each friend and this is multiplied by 7 to obtain this first expression, i.e. 7(2g + 20t –3).

b. The second expression 14g + 140t – 21

The simple interpretation of this expression us that deducting $21 from the to total bill which is 14 games of golf plus 140 tokens is equal to the total amount this group spent.

Therefore, this second expression is obtained by simply opening the bracket in the first expression by multiplying 7 by each of the what is in the bracket.

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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
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Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

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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.

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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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Step-by-step explanation:

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