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2 years ago

A . 12 B . 15 C . 36 D .39

Mathematics

2 answers:

yulyashka [42]2 years ago

7 0

Answer:

The answer is 36 or c all you need to do is multiply 12 by 3

Gnoma [55]2 years ago

7 0

C-36 because your multiplying all of them by 3

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Solve the equation: -7n = 105 What is the value of n?

kirza4 [7]

Answer:

-15

Step-by-step explanation:

process of elimination.

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2 years ago

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Solve for t. −3t≥39 Enter your answer as an inequality in the box.

Flauer [41]

Answer:

$t\leq -13$

Step-by-step explanation:

Given inequality is $−3t\geq39$

Simplifying:

$-t\geq \frac{39}{3}$

$=>-t\geq13$

Sign of the inequality changes of making both sides negative.

$=>t\leq -13$

Hence the solution for the given inequality is represented by $t\leq -13$

This means that the original inequality will be true only for those values of t which are less than or equal to -13.

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2 years ago

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DONT COMMENT IF YOU DON KNOW

tankabanditka [31]

Answer:

okay so im not sure which is being round so i'll do both

if you are going to round the miles and gallons:

278.2 to 278 because the decimal is below 0.5

19.6 to 20 because the decimal is above 0.5

278/20 = 13.9 which you would round to 14 because the decimal is above 0.5

if you are only going to round the answer:

278.2/19.6 = 14.193877551 which you would round to 14 because the decimal is below 0.5

so regardless of what you do,  your answer will be 14

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2 years ago

find the probability of being delt 5 clubs and 3 cards with one of each remaining suit in 8 card poker

kumpel [21]

Answer: 0.003757(approx).

Step-by-step explanation:

Total number of combinations of selecting r things out of n things is given by:-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Total cards in a deck = 52

Total number of ways of choosing 8 cards out of 52 = $^{52}C_8$

Total number of ways to choose 5 clubs and 3 cards with one of each remaining suit = $^{13}C_5\times^{13}C_1\times^{13}C_1\times^{13}C_1$ [since 1 suit has 13 cards]

The required probability = $=\dfrac{^{13}C_5\times^{13}C_1\times^{13}C_1\times^{13}C_1}{^{52}C_8}$

$=\dfrac{\dfrac{13!}{5!8!}\times13\times13\times13}{\dfrac{52!}{8!44!}}\\\\=\dfrac{24167}{6431950}\\\\\approx0.003757$

Hence, the required probability is 0.003757 (approx).

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2 years ago

Help Please thanks!!!!

expeople1 [14]

6 i believe but im not sure

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2 years ago