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BARSIC [14]
3 years ago
13

Nancy sells handcrafted bracelets at a flea market for $7. If her monthly fixed costs are $775 and each bracelet costs her $2.75

to make, how many bracelets must she sell in a month to make a profit?
Mathematics
1 answer:
sleet_krkn [62]3 years ago
8 0
Each bracelet costs $2.75 She sells them for $7 She makes 7-2.75 = $4.25/bracelet   Let b = # bracelets To make a profit she has to earn more than $775   4.25b > 775 b > 775/4.25 b > 182 6/17
  To make a profit she must sell at least 182 bracelets  
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nevsk [136]

It is a cubic function.

8 0
1 year ago
Suppose that a local TV station conducts a survey of a random sample of 120 registered voters in order to predict the winner of
forsale [732]

Answer:

a) The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b) The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

Sample of 120 registered voters in order to predict the winner of a local election. The Democrat candidate was favored by 62 of the respondents.

So 120 - 62 = 58 favored the Republican candidate, so:

n = 120, \pi = \frac{58}{120} = 0.4833

99% confidence level

So \alpha = 0.01, z is the value of Z that has a p-value of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 - 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.3658

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 + 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.6001

The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b. If a candidate needs a simple majority of the votes to win the election, can the Republican candidate be confident of victory? Justify your response with an appropriate statistical argument.

The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

8 0
2 years ago
A fair coin is flipped 4 times. what is the probability that both heads and tails occur?
MatroZZZ [7]
There are 5 possibilities, and they are// H,H,H,H - H,H,H,T - H,H,T,T - H,T,T,T - T,T,T,T// 
4 0
2 years ago
a brownie recipe calls for 1 cup of sugar and 1/2 cup of flour to make one batch of brownies. To make multiple batches, the equa
Lady_Fox [76]

Answer:

A

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f = 1/2s

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This point must be on the graph

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5 0
2 years ago
One thousand tonnes (1000 t, one t equals 10 cubed kg) of sand contains about a trillion (10 super 12) grains of sand. How many
Phantasy [73]

Answer with Step-by-step explanation:

Since 1 mole of sand will contain Avagadro's Number of sand particles (by definition of 1 mole)

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Now since it is given that mass of 1 trillion sand particles is 1000 tonnes Thus the mass of 6.022\times 10^{11} trillion sand particles is

Mass=1000\times 6.022\times 10^{11}=6.022\times 10^{14}tonnes

Part 2)

Since it is given that volume of 1 sand particle is 1.0mm^{3} thus the volume of 1 mole of sand is volume of 6.022\times 10^{23} sand particles

Thus volume of 1 mole is V=1.00\times 10^{-18}km^{3}\times 6.022\times 10^{23}=6.022\times 10^{5}km^{3}

Now since the Area of united states is A=3.6\times 10^{6}mile^{2}=5.8\times 10^{6}km^{2}

Thus the depth of the sand pile is

Depth=\frac{Volume}{Area}=\frac{6.022\times 10^5}{5.8\times 10^6}=0.10387km=103.8meters

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