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3 years ago

MARKING BRAINIEST ANSWER!!!

Mathematics

2 answers:

cluponka [151]3 years ago

3 0

Answer:

y = 3

Step-by-step explanation:

Hello,

$f(x)=(\dfrac{1}{2})^x+3$

when x tends to $+\infty$

$(1/2)^x$ tends to 0 as 1/2 < 1

so the equation of the asymptote for this function is y = 3

Nataliya [291]3 years ago

3 0

Y=3 is the answer. Hope this helps.

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What is the conditional probability that exactly four heads appear when a fair coin is flipped five times

cestrela7 [59]

Answer:

5/32

Step-by-step explanation:

If a coin if flipped five times with four heads, then it has exactly one tail. The number of way to arrange four heads and one tail are as follows:

T H H H H

H T H H H

H H T H H

H H H T H

H H H H T

Essentially, there are five way because there are five "positions" where the tail could be.

Then, you need to find the total amount of possibilities for flipping a coin five times. Every time you flip it, there are two possibilities- heads and tails. Therefore, for five flips, the total amount of possibilities is

2x2x2x2x2 = 32

You know that

probability= # of desired outcome

----------------------------------

# of total outcomes

So the probablity is 5/32.

8 0

2 years ago

Help help help help help help help help help help

AysviL [449]

Answer:

1/4

GO GOOD LUCK

Step-by-step explanation:

U doing IXL? LOL

make me brainliest please

7 0

2 years ago

Read 2 more answers

What is 50/12 in a mixed number

worty [1.4K]

Just divide 50 by 12

4 1/6

8 0

3 years ago

Read 2 more answers

Complete the table for the given rule.

LekaFEV [45]

Answer:

uh 52

Step-by-step explanation:

I say 52 cause like it sounds about reasonable

5 0

3 years ago

Eric wants to get an output of 0. Can he do this with each machine? If so, how? If not, why not?

natita [175]

Answer:

No, he can get an output of 0 with the second machine (function B) but he cannot get an output of 0 with the first machine (function A).

Explanation

The way each machine works is given by the expression (function) inside it.

1) Function A

To get an output of 0 with the function y = x² + 3, you must solve the equation x² + 3 = 0.

Since x² is zero or positive for any real number, x² + 3 will never be less than 3 (the minimum value of x² + 3 is 3). So, it is not possible to get an output of 0 with the first machine.

2) Function B

Solve $y=\sqrt{x}-2=0$

$\sqrt{x}-2=0$

$\sqrt{x} =2$

$x=2^{2}$

$x=4$

So, he can get an output of 0 by using x = 4.

4 0

2 years ago