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cluponka [151]
3 years ago
12

Eman is planning to sell wind chimes at a craft fair. The cost of her tools for building the wind chimes is 130. The cost of mat

erials are $10 per wind chime. What is the total cost to make 100 wind chimes
Mathematics
1 answer:
antoniya [11.8K]3 years ago
7 0

We are given : The cost of tools for building wind chimes = $130.

Cost of material for each wind chime = $10.

We need to find the total cost of making 100 wind chimes.

Please note : $130 is the fix cost and $10 is the variable cost of each wind chime.

<em>In order to find the cost of 100 wind chimes, we need to multiply 100 by cost of making one wind chime and add the fix cost for tools for building wind chimes.</em>

Therefore, total cost of making 100 wind chimes = 10×100 + 130

= 1000+130

=1130.

<h3>Therefore, the total cost to make 100 wind chimes is $1130.</h3>

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Alborosie
3x4=12
12x5=60
Reya played the guitar for 60 hr.
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3 years ago
An automobile manufacturer who wishes to advertise that one of its models achieves 30 mpg (miles per gallon) decides to carry ou
Katyanochek1 [597]

Answer:

t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

p_v =P(t_{(5)}  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation  

The mean and sample deviation can be calculated from the following formulas:

\bar X =\frac{\sum_{i=1}^n x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}

\bar X=29.35 represent the sample mean  

s=1.365 represent the sample standard deviation  

n=6 sample size  

\mu_o =30 represent the value that we want to test  

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

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is lower than 30 or no, the system of hypothesis are :  

Null hypothesis:\mu \geq 30  

Alternative hypothesis:\mu < 30  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

P-value  

We need to calculate the degrees of freedom first given by:  

df=n-1=6-1=5  

Since is a one-side left tailed test the p value would given by:  

p_v =P(t_{(5)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

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<span>Twelve diminished by six times a number</span>

<span> </span>

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