Round 1.9383 to 1 decimal place

A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?

alexandr1967 [171]

Answer:

The probability that the last card dealt is an ace is $\frac{1}{13}$ .

Step-by-step explanation:

Given : A deck of ordinary cards is shuffled and 13 cards are dealt.

To find : What is the probability that the last card dealt is an ace?

Solution :

There are total 52 cards.

The total arrangement of cards is 52!.

There is 4 ace cards in total.

Arrangement for containing ace as the 13th card is $4\times 51!$ .

The probability that the last card dealt is an ace is

$P=\frac{4\times 51!}{52!}$

$P=\frac{4\times 51!}{52\times 51!}$

$P=\frac{4}{52}$

$P=\frac{1}{13}$

Therefore, the probability that the last card dealt is an ace is $\frac{1}{13}$ .

4 0

3 years ago

A drawer contains 12 brown socks and 12 black socks, all unmatched. A man takes socks out at random in the dark. 21. Required in

avanturin [10]

Answer:

6 socks

Step-by-step explanation:

What we must do is calculate the probability of this happening, that he takes out two black socks in the first two taken out.

There are 12 black socks and in total they are 24, therefore the probability of drawing 1 is:

12/24

and now the probability of getting another one is 11 (there is one less outside) and in total they are 23:

11/23

the final probability is the multiplication of these events:

(12/24) * (11/23)

P = 0.24

Now, to know how many you should get, we multiply the probability by the total number of socks, that is:

0.24 * 24 = 5.76

So you must take out at least 6 socks for the above to happen.

6 0

4 years ago

David wants to purchase a new fish tank. His current tank is 36 in. long, 12 in. wide, and 18 in. high. His new tank

Serhud [2]

Answer:

It should be 14400 square inches. Sorry if i'm wrong

3 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago

Find the area of the following shape. (8 points)<br><br> what is the answer?

Ivenika [448]

Answer:

72 square units

Step-by-step explanation:

Identify the height
Identify the length
Multiply those two together
That is your answer

H = 9 units

L = 8 units

A = H × L

A = 9 × 8

A = 72 square units

8 0

2 years ago