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Dimas [21]
3 years ago
10

in 2016 the members of a club paid a total of 42 in subscription fees.In 2017,the number of members were increased by 20 and the

subscription fees were subsequently reduced by 10 cents per member.the total of the subscription fees for 2017 was 45.How many members were there in 2016?
Mathematics
1 answer:
lianna [129]3 years ago
3 0

Answer:

70 members in 2016

Step-by-step explanation:

In 2016

Let the number of members= x

They all paid y amount

Total fees= 42

Xy= 42...... equation 1

In 2017

Members= x+20

They paid y-0.1

Total fees = 45

(X+20)(y-0.1)= 45…equation 2

Xy -0.1x +20y -2= 45

Using values from equation 1

42-0.1(42/y) +20y -2= 45

-4.2/y+20y-5= 0

20y² -4.2 -5y = 0

100y² -25y -21= 0

Solving quadratically

Y= 0.6 or -0.35

We are dealing with price so y Is

definitely a positive number

Y= 0.6

If y= 0.6

Solving for x

Xy= 42

X= 42/0.6

X=70

There were exactly 70 members in 2016

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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

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)

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-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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