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3 years ago

5

What is the answer ????

1 answer:

zlopas [31]3 years ago

7 0

I think its c hope this helps sorry if it dont

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We are given a coin for which the probability of heads is p (0 < p < 1) and the probability of tails is 1 - p. consider a

ra1l [238]

What are the Answer Choices?

6 0

4 years ago

Bradley is purchasing a new car for 37000. If the car loses 20%of its value every year, how much will the car be worth in 7 year

spayn [35]

ANSWER

After 7 years the car will worth, 7,759.5

EXPLANATION

The initial value of the car is 37000.

The rate of depreciation is 20%

In 7 years time, we can calculate the value of the car using the formula,

$A=P(1-r\%)^n$

We substitute the values into the formula to obtain,

$A=37000(1-0.2)^7$

$A=37000(0.88)^7$

$A=7759.5$ to the nearest tenth

6 0

3 years ago

The doubling period of a bacterial population is 20 20 minutes. At time t = 100 t=100 minutes, the bacterial population was 9000

Nuetrik [128]

Answer:

The initial population was 2810

The bacterial population after 5 hours will be 92335548

Step-by-step explanation:

The bacterial population growth formula is:

$P = P_0 \times e^{rt}$

where P is the population after time t, $P_0$ is the starting population, i.e. when t = 0, r is the rate of growth in % and t is time in hours

Data: The doubling period of a bacterial population is 20 minutes (1/3 hour). Replacing this information in the formula we get:

$2 P_0 = P_0 \times e^{r 1/3}$

$2 = e^{r \; 1/3}$

$ln 2 = r \; 1/3$

$ln 2 \times 3 = r$

$2.08 \% = r$

Data: At time t = 100 minutes (5/3 hours), the bacterial population was 90000. Replacing this information in the formula we get:

$90000 = P_0 \times e^{2.08 \; 5/3}$

$\frac{9000}{e^{2.08 \; 5/3}} = P_0$

$2810 = P_0$

Data: the initial population got above and t = 5 hours. Replacing this information in the formula we get:

$P = 2810 \times e^{2.08 \; 5}$

$P = 92335548$

3 0

3 years ago

PLEASE HELP ME!!!<br><br> Which one is bigger 67 1/2 or 112 3/6?

MatroZZZ [7]

112 3/6 im guessing

8 0

3 years ago

If the probability of selecting, without replacement, 2 red marbles from a bag containing only red and blue marbles is 3 55 and

Marat540 [252]

Answer:

Step-by-step explanation:

Let T = the TOTAL number of marbles in the bag.

We are told that 3 of those marbles are red

P(select a red marble first) = 3/T

Once we've removed the first red marble, there are T-1 marbles remaining and 2 of them are red

So, P(select a red marble second) = 2/(T-1)

Okay, now let's use probability rules to answer the question....

P(select 2 red marbles) = P(select a red marble first AND select a red marble second)

= P(select a red marble first) x P(select a red marble second)

= 3/T x 2/(T-1)

= 6/(T² - T)

We're told that P(select 2 red marbles) = 3/55

So, we can write: 6/(T² - T) = 3/55

Cross multiply to get: 3(T² - T) = (6)(55)

Divide both sides by 3 to get: T² - T = (2)(55)

Evaluate: T² - T = 110

Rearrange: T² - T - 110 = 0

Factor: (T - 11)(T + 10) = 0

So, EITHER T = 11 OR T = -10

Since T cannot be negative, it must be the case that T = 11

6 0

3 years ago