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MrMuchimi
2 years ago
15

The financial planner for a beauty products manufacturer develops the system of equations below to determine how many combs must

be sold to generate a profit. the linear equation models the income, in dollars, from selling x plastic combs; the quadratic equation models the cost, in dollars, to produce x plastic combs. according to the model, for what price is each comb being sold?
Mathematics
1 answer:
xxTIMURxx [149]2 years ago
4 0

The income that's made based on the equation when 1 comb is sold is $0.5.

<h3>How to calculate the price?</h3>

From the information given, the equation is given as y = x/2. In this case, it was illustrated that the income is depicted based on the combs sold.

Therefore, the income by selling 1 comb will be:

y = 1/2

y = $0.5

Learn more about equations on:

brainly.com/question/2972832

#SPJ1

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Fiesta28 [93]
The chef can make 25 cakes

Explaintion:
There is a few ways to solve this but an app say way is to
Multiply the 15 gallons by the denominator of the fraction then divide by 3
So 15x5=75
75/3=25
Therefore the chef can make 25 cakes
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Answer:

295.68

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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What is the Area of this triangle? WILL MARK BRAINLIEST
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4 0
2 years ago
A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar
Arisa [49]

Answer:

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

n=17 represent the sample selected

\alpha=0.1 represent the significance

s^2 =4.7^2 =22.09 represent the sample variance

\sigma^2_0 =3.5^2 =12.15 represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 = 12.15

Alternative hypothesis: \sigma^2 \neq 12.15

The statistic for this case is given by:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

The degrees of freedom are:

df =n-1= 17-1=16

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0
3 years ago
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