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3 years ago

Tell me a joke, who ever's joke I like better will get a Brainlist

Mathematics

2 answers:

morpeh [17]3 years ago

7 0

Here’s a joke

What did the baby mushroom say to the mom mushroom?

I don’t have mush-room
(much room)

Me:

FromTheMoon [43]3 years ago

6 0

I wanted to be king because I saw dream of king

one day when I visit to dentist

Dentist: "You need a crown." -

Patient: "Finally someone who understands me"

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Laura creates a rectangular prism with wooden cubes. Each cube has an edge length of 3/4 inch. she uses a total of 240 cubes. th

oksano4ka [1.4K]

2/4 would be the fraction

3 0

3 years ago

Help please anyone! Brainliest reward it counting on it!

Viefleur [7K]

X1 + x2 / 2 , y1 + y2 / 2

8 + x2 /2 = 6

So x is 4

9 + y2 /2 = 6

So y is 3.

Answer is B ( 4, 3)

Please Mark Brainliest if it helped

4 0

3 years ago

Evaluate y/2x-z x=4y=3and z=2

Gwar [14]

First you plug in the missing values
y=3 x=4 z=2
3÷2×4-2
than you solve
the answer is 4

5 0

4 years ago

Suppose a certain computer virus can enter a system through an email or through a webpage. There is a 40% chance of receiving th

DedPeter [7]

Answer:

P = 0.42

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, that is the probability of receiving this virus through the email.

-Set B, that is the probability of receiving it through the webpage.

The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that $A \cap B$ = 0.17.

By email only

The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:

$A + (A \cap B) = 0.40$

$A + 0.17 = 0.40$

$A = 0.23$

where A is the probability that the system receives the virus just through the email.

The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.

By webpage only

There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:

$B + (A \cap B) = 0.35$

$B + 0.17 = 0.35$

$B = 0.18$

where B is the probability that the system receives the virus just through the webpage.

The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.

What is the probability that the virus does not enter the system at all?

So, we have the following probabilities.

- The virus does not enter the system: P

- The virus enters the system just by email: 23% = 0.23

- The virus enters the system just by webpage: 18% = 0.18

- The virus enters the system both by email and by the webpage: 17% = 0.17.

The sum of the probabilities is 100% = 1. So:

P + 0.23 + 0.18 + 0.17 = 1

P = 1 - 0.58

P = 0.42

There is a probability of 42% that the virus does not enter the system at all.

5 0

3 years ago

AYOOO PLZ HELP ASAP!!!

tamaranim1 [39]

Answer:

Step-by-step explanation:

Well we know that

$224=2^{5} *7$

so we can get the 2 outside of the radical

$x^{11} =(x^{5} )^{2} *x$

and we can get the x^2 outside too.

$y^8=y^5*y^3$

and we also can get y outside.

so we have:

$2x^{2}y\sqrt[5]{7xy^3}$

7 0

3 years ago