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3 years ago

Help asap mark you BRAINLY 5 star and heart tysm!

Mathematics

2 answers:

pogonyaev3 years ago

8 0

To find this just do the opposite of what it wants you t do which is multiplication so do division. -27 divided by -3 which is 9 because a negative divided by a negative is a positive.

Hope I helped
Brainliest if satisfied :D

skad [1K]3 years ago

4 0

-3d=-27
The answer is d = 9
Divide -27/9 = -3.

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RSB [31]

Answer: (b)

x g(x)

1 -16

2 -12

3 -8

If g(x) is 4*f(x), then we can find g(x) by multiplying 4 by x-5

g(x) = 4(x-5)

= 4x-20

Now we can plug in 1,2, and 3 for x to see which table makes sense.

g(1) = 4(1) - 20

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g(2) = 4(2) - 20

= -12

g(3) = 4(3) - 20

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Read 2 more answers

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

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