What is 2 + 2? (joke)

Mathematics

2 answers:

andreev551 [17]3 years ago

7 0

Answer:

Baby 2

Step-by-step explanation:

Andrew [12]3 years ago

6 0

Answer:

the answer is 4

Step-by-step explanation:

■■ + ●● = ■■●●

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A certain electronics manufacturer found that the average cost C C to produce x DVD/Blu- ray players can be found using the equa

Kisachek [45]

Answer:

83,491.67 dollars

Step-by-step explanation:

Given that a certain electronics manufacturer ound that the average cost C C to produce x DVD/Blu- ray players can be found using the equation $C=0.03x^2-5x+700$

To find the minimum average cost per machine we can use derivative test.

we find first and second derivative.

If for a particular value of x first derivative is 0 and second derivative is positive, then that x satisfies the minimum cost criteria.

$C=0.03x^2-5x+700\\C'(x) = 0.06x-5\\C"(x) = 0.06>0$

Second derivative is always positive

I derivative =0 gives

$x=5/0.06 \\=83.33\\~83$

For 83 players the cost would be minimum

Minimum cost = $0.03(83)^2-5(83)+700\\=491.67$

3 0

4 years ago

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

nasty-shy [4]

Answer:

(9.6, 25.7) is a 80% confidence interval for the average net change in a student's score after completing the course.

Step-by-step explanation:

We have n = 6, $\bar{x} = 17.6667$ and s = 13.3367. The confidence interval is given by

$\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the $\alpha/2$ th quantile of the t distribution with n-1=5 degrees of freedom. As we want the 80% confidence interval, we have that $\alpha = 0.2$ and the confidence interval is $17.6667\pm t_{0.1}(\frac{13.3367}{\sqrt{6}})$ where $t_{0.1}$ is the 10th quantile of the t distribution with 5 df, i.e., $t_{0.1} = -1.4759$ . Then, we have $17.6667\pm (1.4759)(\frac{13.3367}{\sqrt{6}})$ and the 80% confidence interval is given by (9.6, 25.7)