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2 years ago

The cost per hour of running an assembly line in a

Mathematics

1 answer:

nlexa [21]2 years ago

5 0

Answer:

the minimum production level is costing $800 (0.8×$1000) per hour for 2000 (2×1000) items produced per hour.

Step-by-step explanation:

if there is no mistake in the problem description, I read the following function :

C(x) = y = 0.3x² - 1.2x + 2

I don't know if you learned this already, but to find the extreme values of a function you need to build the first derivative of the function y' and find its solutions for y'=0.

the first derivative of C(x) is

0.6x - 1.2 = y'

0.6x - 1.2 = 0

0.6x = 1.2

x = 2

C(2) = 0.3×2² - 1.2×2 + 2 = 0.3×4 - 2.4 + 2 = 1.2-2.4+2 = 0.8

so, the minimum production level is costing $800 (0.8×$1000) per hour for 2000 (2×1000) items produced per hour.

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A length of a pen is around three inches and is reasonable but all of the others are way too big and can be measured in feet or miles.

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The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from

Vinvika [58]

Answer:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

$df=n_1 +n_2 -2=13+19-2=30$

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

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

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

4 0

3 years ago