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3 years ago

What do graphs of functions have in common?

Mathematics

1 answer:

Sedaia [141]3 years ago

8 0

Answer:

x and y points on a plane

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Fill in the table with whole numbers to make 34 in four different ways

Keith_Richards [23]

It would be:

Tens | Ones
3 + 4
2 + 14
1 + 24
0 + 34

Hope this helped! c:

7 0

3 years ago

Read 2 more answers

How many of each type of Angle the shape has

mel-nik [20]

There would be 4 greater than a right and none are less then a right, they're either obtuse angles, or right

4 0

3 years ago

Read 2 more answers

What number is not part of the domain of f(x)= 3/ x+5<br> (3 over x + 5)

nignag [31]

-5, you never want the denominator to equal zero or else it doesnt exist!

x+5=0
x=-5

7 0

3 years ago

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:

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

4 0

3 years ago

Please help I will give you brainliest

mezya [45]

Answer:

Answer is 5 3/4

Step-by-step explanation:

Thanks Hope this helps

5 0

2 years ago