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Sergio [31]
3 years ago
15

The circulation of a city library for September was 157,928 books; for October, 184,211 books. What was the percent of gain for

October over September?
Mathematics
1 answer:
ASHA 777 [7]3 years ago
4 0

Answer:

17.0 %

Step-by-step explanation:

% Gain = Difference/Original value × 100 %

Data:

September = 157 928 books

   October =  184 211   books

Calculations:

  Difference =  October - September

                     = 184 211 - 157 928

                     = 26 913

% Difference = 26 913/157 928 × 100 %

                     = 17.0 %

The percent gain for October over September was 17.0 %.

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f(3) = 8

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f(x) = 5x - 7

Put x as 3 and evaluate.

f(3) = 5(3) - 7

f(3) = 15 - 7

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Maureen pought 3 pairs of shoes that cost $37.98 each.If the tax rate is 8%, how much did the Maureen pay?"
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Blake’s Blacksmith Co. produces two types of shotguns, a 12-gauge and 20-gauge. The shotguns are made through a joint production
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Answer:

  $1800 of joint production costs are allocated to each type of shotgun. That is $60 per shotgun.

Step-by-step explanation:

Equal numbers of shotguns of each type are produced in the joint process, so the joint process cost is allocated equally between the types. The allocation for each shotgun is ...

  $3600/(30+30) = $60 . . . . allocated cost per shotgun of each type

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3 years ago
The magic dragon cigarette company claims that their cigarettes contain an average of only 10 mg of tar. a random sample of 100
9966 [12]

Answer:

t=\frac{11.5-10}{\frac{4.5}{\sqrt{100}}}=3.333    

Critical value

The significance is 5% so then \alpha=0.05 and \alpha/2=0.025 then the critical value for this case would be z_{\alpha/2}= 1.64. Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true mean is higher than 10 mg

P value

The p value would be given

p_v =P(z>3.333)=0.000434  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 10 mg

Step-by-step explanation:

Information given

\bar X=11.5 represent the sample mean

\sigma=4.5 represent the population standard deviation

n=100 sample size  

\mu_o =10 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is higher than 10mg, the system of hypothesis would be:  

Null hypothesis:\mu \leq 10  

Alternative hypothesis:\mu > 10  

the statistic for this case would be given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing we got:

t=\frac{11.5-10}{\frac{4.5}{\sqrt{100}}}=3.333    

Critical value

The significance is 5% so then \alpha=0.05 and \alpha/2=0.025 then the critical value for this case would be z_{\alpha/2}= 1.64. Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true mean is higher than 10 mg

P value

The p value would be given

p_v =P(z>3.333)=0.000434  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 10 mg

6 0
2 years ago
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