What is 200 percent of (0.020(5/4) + 3 ((1/5)

A video game sells at Arnolds for $14.99. Arnolds marks the game up at 40% of the selling price. The cost of the video game to A

SVETLANKA909090 [29]

$6.00 for the video game

7 0

2 years ago

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A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter

faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let c (t ) denote the amount of chlorine (in grams) in the tank at time t .

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(c (t )/(1000 + (10 - 25)t ) g/L) * (25 L/s) = 5c (t ) /(200 - 3t ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after t s there will be (1000 + 10t ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time t is (1000 - 15t ) L. Then the concentration of chlorine per unit volume is c (t ) divided by this volume.

So the amount of chlorine in the tank changes according to

$\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}$

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5t )^(5/3), then integrate both sides to solve for c (t ):

$\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0$

$\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0$

$\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0$

$\dfrac{c(t)}{(200-3t)^{5/3}}=C$

$c(t)=C(200-3t)^{5/3}$

There are 20 g of chlorine at the start, so c (0) = 20. Use this to solve for C :

$20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}$

$\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}$

7 0

1 year ago

On a trip, you had to change your money from dollars to euros. You got 450 euros for 600 dollars. Which is a unit rate that desc

chubhunter [2.5K]

The number of Euros received against 600 dollars = 450
Then
For 1 dollar, the number of euros = (450/600) euros
   = (45/60) euros
   = ( 3/4) euros
   = 0.75 euros
So the unit rate that describes the exchange rate is 0.75 euros per dollar. So the correct option among all the options given in the question is option "A". I hope the procedure for getting to the correct result is clear enough for you to understand. In future you can use this procedure for solving similar problems.

7 0

2 years ago

It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are giv

bonufazy [111]

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.9

p-value = 0.206

Since the p-value is greater than α therefore, we cannot reject the null hypothesis.

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

Step-by-step explanation:

Let σ₁² denotes the variance of night shift-workers

Let σ₂² denotes the variance of day shift-workers

State the null and alternative hypotheses:

The null hypothesis assumes that the variance of night shift-workers is equal to or less than day-shift workers.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance of night shift-workers is more than day-shift workers.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic or also called F-value is calculated using

Test statistic = Larger sample variance/Smaller sample variance

The larger sample variance is σ₁² = 38

The smaller sample variance is σ₂² = 20

Test statistic = σ₁²/σ₂²

Test statistic = 38/20

Test statistic = 1.9

p-value:

The degree of freedom corresponding to night shift workers is given by

df₁ = n - 1

df₁ = 9 - 1

df₁ = 8

The degree of freedom corresponding to day shift workers is given by

df₂ = n - 1

df₂ = 8 - 1

df₂ = 7

We can find out the p-value using F-table or by using Excel.

Using Excel to find out the p-value,

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.9, 8, 7)

p-value = 0.206

Conclusion:

p-value > α

0.206 > 0.05 ( α = 1 - 0.95 = 0.05)

Since the p-value is greater than α therefore, we cannot reject the null hypothesis corresponding to a confidence level of 95%

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

3 0

2 years ago

It is desired to test H0: μ = 55 against H1: μ < 55 using α = 0.10. The population in question is normally distributed with a

patriot [66]

Answer:

Step-by-step explaation: 42% of 85 is what number

3 0

2 years ago

What is 200 percent of (0.020(5/4) + 3 ((1/5) - (1/4)))