Please help with question. I’m stuck!

Mathematics

1 answer:

Arisa [49]3 years ago

3 0

There would probably be multiple ways to solve this, but I will only show you one of them.
First, we need to figure out how many tablespoons of the chemical solution it takes to decrease the number of bacteria in a gallon to 300. So we subtract 1200 by 300, to find the number of unnecessary bacteria, which is 900. Now we divide 900 by 75, which is 12. So it would take 12 tablespoons of the chemical solution to make a gallon have 300 bacteria.
There are 500 gallons, and we need to put 12 tablespoons of the chem-solution in each gallon. So we multiply 500 by 12, which is 6,000. Meaning that 6,000 tablespoons of chemical solution would be needed to make the 500-gallon tank have 300 bacteria per gallon.

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4 •f(6) - 6•g(5) =<br> please help me

Alborosie

Answer:

24f−30g

Step-by-step explanation:

4 0

3 years ago

The cost of labor for servicing cars at B&B Automotive is $50 for each whole hour or for any fraction of an hour. There is a

Masja [62]

Answer:

Step-by-step explanation:

We can use the equation of a straight line to model the cost of servicing the car.

let the cost be y and the number of hours be x

and the charge per hour is m

y=mx+c

y=50x+25

given that the time is 3.45 hours it is assumed that the charge is for 4 hours since for a fraction of 0.45 hours we are charged $50

y=50(4)+25

y=200+25

y=$225

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4 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$