Ask question

Ask question

All categories

3 years ago

6

Before his basketball game, Jessie is responsible for cleaning the visiting teams team's half of the court. Three of his teammat

es are responsible for cleaning the home teams half of the court. How much of the entire court is each of jessie's teammates responsible for cleaning?
A:3/2
B:1/5
C:2/3
and D:1/6?

1 answer:

Zepler [3.9K]3 years ago

5 0

It is D:1/6 because 3 of them do the home side and he cleans the other side.

You might be interested in

stack of mail consists of 8 bills, 10 letters, and 6 advertisements. One piece of mail is drawn at random and put aside. Then a

kozerog [31]

INFORMATION:

We know that:

- stack of mail consists of 8 bills, 10 letters, and 6 advertisements.

- One piece of mail is drawn at random and put aside. Then a second piece of mail is drawn.

And we must find P (both are letters)

STEP BY STEP EXPLANATION:

To find the probability, we need to know that we have two events. First, when one piece of mail is drawn at random and put aside and, second, when a second piece of mail is drawn.

These two events are dependent. If A and B are dependent events, P(A and B) = P(A) • P(B after A) where P(B after A) is the probability that B occurs after A has occurred.

So, first

- Probability of A (the first piece is letter)

$P(A)=\frac{favorable\text{ }cases}{total\text{ cases}}=\frac{10}{24}$

- Probability of B after A

Since A already occurred and one piece of the mail was drawn (a letter), now in total we would have 9 letter and 23 total pieces

$P(B\text{ after }A)=\frac{9}{23}$

Finally, replacing in the initial formula

$P(A\text{ and }B)=\frac{10}{24}\cdot\frac{9}{23}=\frac{90}{552}=0.1630$

Finally, the probability would be 0.1630

ANSWER:

P (both are letters) = 0.1630

8 0

1 year ago

Megan has a pen for her dog. The dimensions of the pen are in meters (m) as shown below.

adoni [48]

Answer:

22 m²

Step-by-step explanation:

I think jxxjxjcjnfvnjvchhd

4 0

3 years ago

Read 2 more answers

According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the

FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

Then, we obtain two differential equations

$\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q$

where $k_1$ and $k_2$ are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

$\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

Now, the variables are separated, $P$ and $Q$ appear only on the left, and $t$ appears only on the right, so that we can integrate both sides.

$\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

$\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$ ,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

$e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

$P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$ ,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$ .

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

$P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

$C = P(0) = C_1 e^0 \implies C= C_1$

Similarly, we obtain $C = C_2$ and

$P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}$

The relation between the decay constant $k$ and the half-life is given by

$\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$ . Thus,

$4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

$7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

$P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

$\frac{P(t)}{Q(t)} = 137.7$

Hence,

$\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7$

Solving for $t$ yields $t \approx 6 \times 10^9$ , which means that the age of the universe is about 6 billion years.

5 0

3 years ago

Please help me with this ❤️❤️❤️

tester [92]

Answer: h=P/mg

Step-by-step explanation:

7 0

3 years ago

Which set of numbers represents the domain for the following table below

elena-14-01-66 [18.8K]

Answer:

21

Step-by-step explanation:

21

8 0

3 years ago