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Which equations are equivalent to the given equation?

Strike441 [17]

49=-7k is equivalent to D because when you simplify both of them, you get -7=k

6 0

2 years ago

A new car is purchased for 23700 dollars.the value o the car depreciates at 13.5% per year what will be the value of the car be,

lara31 [8.8K]

it will be 31,995 but we have to estimate yo the nearest cent so 10 cents

5 0

3 years ago

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Find the indicated conditional probability

andrezito [222]

Given:

The two way table.

To find:

The conditional probability of P(Drive to school | Senior).

Solution:

The conditional probability is defined as:

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

Using this formula, we get

$P(\text{Drive to school }|\text{ Senior})=\dfrac{P(\text{Drive to school and senior})}{P(\text{Senior})}$ ...(i)

From the given two way table, we get

Drive to school and senior = 25

Senior = 25+5+5

= 35

Total = 2+25+3+13+20+2+25+5+5

= 100

Now,

$P(\text{Drive to school and senior})=\dfrac{25}{100}$

$P(\text{Senior})=\dfrac{35}{100}$

Substituting these values in (i), we get

$P(\text{Drive to school }|\text{ Senior})=\dfrac{\dfrac{25}{100}}{\dfrac{35}{100}}$

$P(\text{Drive to school }|\text{ Senior})=\dfrac{25}{35}$

$P(\text{Drive to school }|\text{ Senior})=0.7142857$

$P(\text{Drive to school }|\text{ Senior})\approx 0.71$

Therefore, the required conditional probability is 0.71.

5 0

3 years ago

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A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

2 years ago

HELP QUICK PLEASE WILL GOVE BRAINLEOCMWdiowemforewmfo

hodyreva [135]

Answer:

Assignment: 01.07 Laboratory TechniquesAssignment: 01.07 Laboratory TechniquesAssignment: 01.07 Laboratory Techniques

Step-by-step explanation:

Assignment: 01.07 Laboratory Techniques

Assignment: 01.07 Laboratory TechniquesAssignment: 01.07 Laboratory Techniques

8 0

1 year ago

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