Ask question

Ask question

All categories

2 years ago

7

A certain game consists of rolling a single fair die and pays off as follows: $7 for a 6, $2 for a 5, $1 for a 4, and no payoff

otherwise.

1 answer:

Natalka [10]2 years ago

8 0

Answer:

E Winning) = $ 1.667

Step-by-step explanation:

E(winnings) = (7 x ( $\frac{1}{6}$ )) + (2 x ( $\frac{1}{6}$ + (1 x ( $\frac{1}{6}$ ))

= $\frac{7}{6}$ + $\frac{2}{6}$ + $\frac{1}{6}$

= $\frac{10}{6}$

= $\frac{5}{3}$

= $ 1.667

You might be interested in

Can someone please help me with #3?

kap26 [50]

Answer:

Step-by-step explanation:

First, make sure you know which section of the number line you want as 1 and 2. Next, put 0.25 not right at 0 but very close, then put 0.75 farther but not too much from 0.25. Then for the decimal 1.99 put that very close to where you marked the 2 but, not on the 2. Finally, put 2.03 very close to 2 but not exactly on the 2. Also, make sure that the number line is marked evenly.

7 0

2 years ago

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

8 0

2 years ago

Two dogs run around a circular track 300 m long. One dog runs at a steady rate of 15m per second, the other at a steady rate of

mihalych1998 [28]

Answer:

300 seconds

Step-by-step explanation:

The first dog run at v₁ = 15 m/sec the second one run at v₂ = 12 m/sec

we know that d = v*t then t = d/v

Then the first dog will take 300/ 12 = 25 seconds to make a turn

The second will take 300 / 15 = 20 seconds to make a turn

Then the first dog in 12 turns 12*25 will be at the start point, and so will the second one at the turn 15.

To check first dog 12 * 25 = 300

And the second dog 15 * 20 = 300

That means that time required for the two dogs to be at the start point together is

300 seconds, in that time the first dog finished the 12 turns, and the second had ended the 15.

Another procedure to solve this problem is as follows:

between 12 m/sec and 15 m/sec the minimum common multiple is 300 ( 300 is the smaller number that accept 12 and 15 as factors 12*15 = 300) Then when time arrives at 300 seconds the two dogs will be again in the starting point

3 0

2 years ago

A bowl has 15 apples, and 10 of the apples are red. To the nearest whole percent, what percent of the apples are red? (thank you

steposvetlana [31]

Answer:

The answer is 66%

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please help!! I will give brainliest thank you!!

mamaluj [8]

$\bold{\huge{\red{\underline{Solution}}}}$

$\bold{\underline{ Given :- }}$

We have given two figures that rectangle and parallelogram
The base and height of parallelogram is 7ft and 5ft
The length and breath of rectangle is 3ft and 1ft .

$\bold{\underline{ To \: Find :- }}$

We have to find the area of shaded part?

$\bold{\underline{ Let's \: Begin :- }}$

Firstly, we will calculate the area of parallelogram then area of rectangle after that we will subtract the area of rectangle from the area of parallelogram then we got the area of shaded part.

$\sf{\pink{Area \:of\: parallelogram = Base * height }}$

$\sf { = 7 × 5 }$

$\sf { = 35ft }$

$\sf{ Thus, \:The \:area\: of \:parallelogram\: is \:35ft}$

$\sf{\underline{ Now}}$

$\sf{\blue{Area\: of\: rectangle = L × B }}$

$\sf{ = 3 × 1 }$

$\sf{ = 3ft}$

$\sf{Thus, \:The\: Area\: of \:rectangle\: is\: 3ft}$

$\sf{\underline{ Now}}$

$\sf{ Total\: area \:of \:shaded\: part }$

$\sf{ Area\: of\: parallelogram - Area\: of\: rectangle }$

$\sf{ = 35 - 3}$

$\sf{ = 32 ft}$

$\bold{\pink{Hence, \: Area\: of \:shaded\: part \:is\: 32ft}}$

8 0

1 year ago