Someone please help me, I need the type of statement and reason.

Here are two rectangles.

jolli1 [7]

Answer:

1/1/3 is the answer “ one and one third

Step-by-step explanation:

to find length scale factor,

take the length of the 2nd triangle and divide it by the length of the first triangle,

you'll get the length scale factor

6 0

3 years ago

Read 2 more answers

Can someone please help me with problem number 13. 14 ASAP

Tanya [424]

Answer:

For 13: 13/30

Step-by-step explanation:

First simplify!

9x - 25 + 21x - 35 + 12 = 0

then simplify some more

9x + 21x = 30x

(-25) + (-35) + 21 + 12 = -13

30x - 13 = 0

now add 13 to both sides, 13 and -13 cancel each other out and it becomes

30x = 13

then you divide by 30 to get x by itself. 30 cancels itself out and you are left with 13/30 which doesn't simplify so that shod be your answer! I couldn't figure out 14 but I hope this helps!!!

4 0

3 years ago

A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability