John and Alan have a collection of x baseball cards. John has x/4 cards. What fraction of the cards does Alan have?

Using what you know about similar figure, solve for X

ale4655 [162]

Since the scale factor is 3:1 you can multiply the length of the side of the triangle to the right by 3 to make the ratio 3:3

the equation would be:
2x+5 = 45

subtract 5 from both sides
2x = 40

divide both sides by 2

x = 20

hope this helps

3 0

3 years ago

There are 40 children in the park which is 500 square yards, and there are 30 children in the baseball field which is 300 square

Karolina [17]

Answer:

30 in 300 is more

Step-by-step explanation:

5 0

3 years ago

Two events E1 and E2 are called independent if p(E1 â© E2) = p(E1)p(E2). For each of the following pairs of events, which are su

posledela

Answer: a) Independent

b) Independent

c) Dependent

Step-by-step explanation:

Since, If a coin is tossed three times,

Then, total number of outcomes, n(S) = 8

a) $E_1$ : tails comes up with the coin is tossed the first time;

$E_1$ = { TTT, THH, THT, TTH }

$E_2$ : heads comes up when the coin is tossed the second time.

$E_2$ = { THT, HHH, THH, HHT }

Thus, $n(E_1)=4$

⇒ $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly, $P(E_2)=\frac{1}{2}$

⇒ $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Since, $E_1\cap E_2$ = { THH, THT }

$n(E_1\cap E_2) = 2$

⇒ $P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{2}{8}=\frac{1}{4}$

Thus, $P(E_1\cap E_2)=P(E_1)\timesP(E_2)$

Therefore, $E_1$ and $E_2$ are independent events.

B) $E_1$ : the first coin comes up tails

$E_1$ = { TTT, THH, THT, TTH }

$E_2$ : two, and not three, heads come up in a row

$E_2$ = { HHT, THH }

Thus, $n(E_1)=4$

⇒ $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly, $P(E_2)=\frac{1}{4}$

⇒ $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

Since, $E_1\cap E_2$ = { THH }

$n(E_1\cap E_2) = 1$

⇒ $P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{1}{8}$

Thus, $P(E_1\cap E_2)=P(E_1)\timesP(E_2)$

Therefore, $E_1$ and $E_2$ are independent events.

C) $E_1$ : the second coin comes up tails;

$E_1$ = { HTH, HTT, TTT, TTH }

$E_2$ : two, and not three, heads come up in a row

$E_2$ = { HHT, THH }

Thus, $n(E_1)=4$

⇒ $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly, $P(E_2)=\frac{1}{4}$

⇒ $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

Since, $E_1\cap E_2$ = $\phi$

$n(E_1\cap E_2) = 0$

⇒ $P(E_1\cap E_2) = 0$

Thus, $P(E_1\cap E_2)\neq P(E_1)\timesP(E_2)$

Therefore, $E_1$ and $E_2$ are dependent events.

3 0

3 years ago

in a class of 24 students, seven eigths are allowed to attend recess on friday. how many students will attend recess

Verizon [17]

Seven eighth = 7/8
8 (Denominator) times 3 = 24
So, you would multiply 7 (numerator) by 3 as you do with the denominator 8.
7 x 3 = 21
21 out of 24 students are allowed to attend recess.
I hope this answer helps you!

8 0

3 years ago

I suck at math so plz help me

Colt1911 [192]

Can u show the graph more

7 0

2 years ago