What is the y intercept of 8x+4y=16

The difference between the MDC and the LDC's

Contact [7]

Answer: MDCs have advanced socially and economically, whereas LDCs are in the early stages of development.

8 0

2 years ago

OMG PLS HELP WITH THIS IM PANICKING OMG I GOT A F IN MATH MY MOM JUST YELLED AT ME IM CRYING PLS HELP WITH THS- PLSS TELL ME WHA

4vir4ik [10]

Answer:

calm down rosie.... it's okay to not be too good at math... take Justin Timberlake. in the show "Friends with Benefits" he's horrible at math but good at life. so don't worry too much. You are still a good person.. even if you're bad at math :P

Step-by-step explanation:

now for the math part

as below

3 0

3 years ago

A ladder is leaning against a wall that is 20 ft tall. The base of the ladder is 30 ft from the wall. How long is the ladder?

creativ13 [48]

15 ft. so bruh i hope this helps you

3 0

3 years ago

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Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h

Delvig [45]

Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

$T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182$

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

$T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148$

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

$T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378$

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

$T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899$

So 2899 possible ways.

4 0

3 years ago

PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE HELP ME!!!!!!!! THATS ALL. I WILL GIVE POINTS, FAN, MEDAL. PLEASE, I AM SO BAD AT MATH

aleksandrvk [35]

"measures to the nearest tenth of a pound" tell us to round each of the weights listed to the nearest tenth (since the unit is already pounds, no conversion is needed). What this means is if the second number to the right of the decimal is 5 or greater, then we add 1 to the first number to the right of the decimal (and replace the second number with a zero). With all this said, it's much easier to show you, rather than to try to explain it. :D

4.35 is rounded up to 4.4 because the second number to the right of the decimal is 5.
Packet A = 4.4
Packet B = 4.4
Packet C = 4.5
Packed D = 4.4
The question is asking what the heaviest reading will be, so from the whole list above we pick the highest number which is 4.5 pounds. Notice how the answer is not Packet C, or 4.48 pounds, but 4.5 pounds. It is asking for the reading, not the packet, nor the weight. Hope this helps.

7 0

3 years ago

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