What is the value of 5 in 5,708

Suppose a tank contains 400 gallons of salt water. If pure water flows into the tank at the rate of 7 gallons per minute and the

Strike441 [17]

Answer:

Step-by-step explanation:

This is a differential equation problem most easily solved with an exponential decay equation of the form

$y=Ce^{kt}$ . We know that the initial amount of salt in the tank is 28 pounds, so

C = 28. Now we just need to find k.

The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is $\frac{dy}{dt}$ . Thus, the change in the concentration of salt is found in

$\frac{dy}{dt}=$ inflow of salt - outflow of salt

Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

$3(\frac{y}{400})$

Therefore,

$\frac{dy}{dt}=0-3(\frac{y}{400})$ or just

$\frac{dy}{dt}=-\frac{3y}{400}$ and in terms of time,

$-\frac{3t}{400}$

Thus, our equation is

$y=28e^{-\frac{3t}{400}$ and filling in 16 for the number of minutes in t:

y = 24.834 pounds of salt

6 0

3 years ago

A study was made of seat belt use among children who were involved in car crashes that caused them to be hospitalized. It was fo

borishaifa [10]

Answer:

a) $P(5$

And we can find this probability with this difference:

$P(-1.90$

And using the norma standard distribution or excel we got:

$P(-1.90$

b) $P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)$

And using the complement rule we got:

$P(Z>-1.096) =1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(7.37,1.25)$

Where $\mu=7.37$ and $\sigma=1.25$

We are interested on this probability

$P(5$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(5$

And we can find this probability with this difference:

$P(-1.90$

And using the norma standard distribution or excel we got:

$P(-1.90$

Part b

For this case we want this probability:

$P(X>6)$

And we can use the z score and we got:

$P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)$

And using the complement rule we got:

$P(Z>-1.096) =1-P(Z$

8 0

3 years ago

A marketing firm would like to test-market the name of a new energy drink targeted at 18- to 29-year-olds via social media. A st

Anon25 [30]

Answer:

(a) The probability that a randomly selected U.S. adult uses social media is 0.35.

(b) The probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c) The probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

Step-by-step explanation:

Denote the events as follows:

X = an US adult who does not uses social media.

Y = an US adult between the ages 18 and 29.

Z = an US adult between the ages 30 and above.

The information provided is:

P (X) = 0.35

P (Z) = 0.78

P (Y ∪ X') = 0.672

(a)

Compute the probability that a randomly selected U.S. adult uses social media as follows:

P (US adult uses social media (X')) = 1 - P (US adult so not use social media)

$=1-P(X)\\=1-0.35\\=0.65$

Thus, the probability that a randomly selected U.S. adult uses social media is 0.35.

(b)

Compute the probability that a randomly selected U.S. adult is aged 18–29 as follows:

P (Adults between 18 - 29 (Y)) = 1 - P (Adults 30 or above)

$=1-P(Z)\\=1-0.78\\=0.22$

Thus, the probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c)

Compute the probability that a randomly selected U.S. adult is 18–29 and a user of social media as follows:

P (Y ∩ X') = P (Y) + P (X') - P (Y ∪ X')

$=0.22+0.65-0.672\\=0.198$

Thus, the probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

6 0

3 years ago

Name the ray in the figure

Pavlova-9 [17]

the answer is m to p

5 0

3 years ago

Read 2 more answers

The product of the ages of two adults is 572. Find their ages.

AlladinOne [14]

Answer:

286

Step-by-step explanation:

7 0

3 years ago