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2 years ago

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Joy makes 6.5 litres of soup, correct to the nearest 0.5 litre. She serves the soup in 280 ml portions, correct to the nearest 1

0ml. 22 people order this soup. Does Joy have enough soup

2 answers:

max2010maxim [7]2 years ago

8 0

Amount of soup made by Joy = 6.5 liters and nearest to 0.5 liters is 7 liters

Amount of soup in each serving portion = 280 ml. As the last digit is zero, this will not be rounded off and nearest value will remain 280

Number of people being served = 22

So total consumption of soup for 22 people is =

$22\times280=6160$ = 6160 ml

In liters this is = $\frac{6160}{1000}=6.160$ liters

Hence, Joy has enough soup for 22 people.

Levart [38]2 years ago

7 0

Answer:

Joy definitely does have enough soup for 22 people.

Step-by-step explanation:

Upper Bound: 285ml turn it to 0.285 L 10÷2=5

280ml :

Lower Bound: 275ml turn it to 0.275 L

Upper Bound: 6.75 L 0.5÷2=0.25

6.5 Litters :

Lower Bound:6.25 L

6.75÷0.275=24.54 or 25

6.25÷0.285=21.92 or 22

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

(-0.0225, -0.0375)

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