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sergey [27]
3 years ago
8

A craft club makes 95 identical paperweights to sell. They collect $230.85 from selling all of the paperweights. If the profit t

he club collects is two times as much as the cost to make each one, what does it cost the club to make each paperweight?
Mathematics
1 answer:
Liula [17]3 years ago
6 0

Answer: It costs $1.22 to make each paperweight.

Step-by-step explanation:

Given : Selling price of 95 identical paperweights = $230.85

Then, Selling price of 1 paperweight = (Selling price of 95 identical paperweights) ÷ 95

= $ (230.85 ÷ 95) = $ 2.43

Now, selling price of each paperweight = 2 x ( cost of each paperweight)

⇒ cost of each paperweight = (selling price of each paperweight ) ÷ 2

⇒ cost of each paperweight = $ 2.43 ÷ 2 = $ 1.215 ≈ $ 1.22

Hence, it costs $1.22 to make each paperweight.

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Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

Population mean, μ = 425 mg/L

Sample mean, \bar{x} = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

Now, t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108

Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0
3 years ago
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