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ANEK [815]
2 years ago
14

HURRY PLEASE SHOW UR WORK AND WILL GIVE BRAINLIEST !!!

Mathematics
2 answers:
Rama09 [41]2 years ago
8 0
Hope this help enjoy ya day
EastWind [94]2 years ago
3 0

Answer:

im not exactly sure, but i think it would be PART A: either option B or D by going diagonal across the grid and then normal across the grid. PART B: option B because that is the shortest option going diagonal that matches with the options available

Step-by-step explanation:

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What terms can be combined with 3a? Select all that apply.
antiseptic1488 [7]

Answer:

You can only add like terms with like terms, so the terms that can be combined with 3a are the terms that end in an a.

So, the answers are:

4a and 14a

Let me know if this helps!

6 0
2 years ago
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Someone please help me with number 9 and 10
NemiM [27]
9 is c-point L
10 is a- $10,000
3 0
3 years ago
Solve<br> -31-4x=-5-5(1+5x)
MrRa [10]
-31 - 4x = 5 - 5 ( 1 + 5x )
-31 - 4x = 5 - 5 - 25x
-31 - 4x = -25x
+ 31 to both sides
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4 0
3 years ago
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What are Three common multiples of 3 and 2?
Akimi4 [234]

Answer:

Common multiples of 2 and 3 include 6, 12, 18, and 24.

Step-by-step explanation:

A common multiple is a number that is a multiple of two or more numbers.

6 0
3 years ago
A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orang
laiz [17]

<u>Answer-</u>

a. Probability that  three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

<u>Solution-</u>

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

\binom{19}{3} = \frac{19!}{3!16!} = 969

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

And this process can be done in,

\binom{33}{6} = \frac{33!}{6!27!} =1107568

So total number of selection = (969)×(1107568) = 1073233392

Drawing 9 candies out of 52 candies,

\binom{52}{9} = \frac{52!}{9!43!} = 3679075400

∴P(3 white candies) = \frac{1073233392}{3679075400} =0.29



Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,

\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}

=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})

=(969)(45)(7)(5)(15)=22892625

Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,

\binom{52}{9}=\frac{52!}{9!43!} =3679075400

∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =

\frac{22892625}{3679075400} = 0.006


6 0
3 years ago
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