Solve this system of equations using substitution:

What 468 times 120 divided by 2

cluponka [151]

Answer:

28080

Step-by-step explanation:

Firstly, multiply 468 with 10, which gets you 4680. Then multiply that with 12 which gets you 56160.

After you take that, divide the number by 2.

Final answer is 28080.

7 0

3 years ago

Read 2 more answers

If C(x) is the cost of producing x units of a commodity, then the average cost per unit is c(x) = C(x)/x. Consider the cost func

nadya68 [22]

Answer:

a) C(1,000)=288,491.11 $

b) c(1,000)=288.49 $/u

c) dC/dx(1,000)=349.74 $/u

d) x=100 u

e) c=220 $/u

Step-by-step explanation:

(a) Find the total cost at a production level of 1000 units.

$C(x) = 2,000 + 160x + 4x^{3/2}\\\\C(1,000)=2,000 + 160(1,000) + 4(1,000)^{3/2}\\\\C(1,000)=2,000+160,000+126,491.11= 288,491.11$

(b) Find the average cost at a production level of 1000 units.

$c(x)=C(x)/x\\\\c(1,000)=C(1,000)/1,000=288,491.11/1,000=288.49$

(c) Find the marginal cost at a production level of 1000 units.

$\frac{dC}{dx} =0+160+4*(3/2)x^{3/2-1}=160+6x^{1/2}\\\\dC/dx|_{1,000}=160+6*(1,000)^{1/2}=160+189.74=349.74$

(d) Find the production level that will minimize the average cost.

$c=2,000x^{-1}+160+4x^{1/2}\\\\dc/dx=2,000(-1)x^{-2}+0+4(1/2)x^{-1/2}=0\\\\-2,000x^{-2}+2x^{-1/2}=0\\\\2x^{-1/2}=2,000x^{-2}\\\\x^{-1/2+2}=1,000\\\\x^{3/2}=1,000\\\\x=1,000^{2/3}=100$

(e) What is the minimum average cost?

$c(x) = 2,000/x + 160 + 4x^{1/2}\\\\C(100)=2,000/100 + 160 + 4(100)^{1/2}=20+160+40=220$

5 0

3 years ago

For two days Winston rescued tadpoles from a puddle he rescued 54 on Friday this is 17 less than the number he rescued on Sunday

Studentka2010 [4]

Answer:

t - 17 = 54

Step-by-step explanation: look it up

3 0

3 years ago

............................

prohojiy [21]

Ahh yes, negative exponents always give us a scare once and a while. All the negative means is to flip the position of its base. For instance, if x has a negative exponent and x in the denominator, all you would have to do is move x to the numerator with the same power (except it's no longer negative). Before we substitute x and all the other variables which the values given, let's eliminate the negatives first.

After flipping positions/eliminating the negative exponents it should look like this:

$\frac{ 2^{2} x^{3} }{ y^{5} }$
which simplifies to
$\frac{ 4 x^{3} }{ y^{5} }$
now that everything is simplified, and all negative exponents are eliminated we can substitute x with 2, and y with (-4).
$\frac{4 (2)^{3} }{ (-4)^{5} }$
which simplifies to
$\frac{4(8)}{(-1024)} = \frac{32}{-1024} = - \frac{1}{32}$

Final Answer: - \frac{1}{32} [/tex]

4 0

3 years ago

Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

n = 20

s = 3.9

Confidence level = 95%

⇒ α = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

4 0

3 years ago