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3 years ago

How do you vibe check

Mathematics

2 answers:

miskamm [114]3 years ago

7 0

Just check the person’s vibe lolz

ziro4ka [17]3 years ago

4 0

Answer:

good question.... maybe by the way a person acts

Step-by-step explanation:

You might be interested in

9 is the difference between two numbers, and their sum is 31. Find the numbers.

algol [13]

Answer:

The first number is 20, the second one is 11

Step-by-step explanation:

Let the first number be x and the second number be y

Since it says "9 is the difference between two numbers", the first equation will be x-y=9

It also says that "their sum is 31". That means the second equation will be x+y=31

here are the two equations:

x-y=9

x+y=31

You can solve this any way you like, but I'll show elimination

add the two equations

x-y=9

x+y=31

2x=40

divide by 2

x=20

Substitute 20 as x in one of the equations

20+y=31

subtract 20 from both sides

y=11

That means the two numbers are 20 and 11

Hope this helps!

4 0

2 years ago

There is a 60% probability that a home in the United States has a large-screen TV. In a sample of six homes, what is the probabi

lesya [120]

Answer:

faster and smarter Homes its b

Step-by-step explanation:

I did this one

3 0

2 years ago

What two numbers have the sum of 12 and the difference of 4

Veronika [31]

Answer:

8 and 4

Step-by-step explanation:

8+4= 12

8-4= 4

hope this helped♥

5 0

2 years ago

Read 2 more answers

Plzzz help me with this

NeX [460]

You would say that the side ADAC is congruent to ABCA because the two triangles where it is split in half are congruent as well.

3 0

2 years ago

One thousand independent rolls of a fair die will be made. Compute an approximation to the probability that the number 6 will ap

pashok25 [27]

Answer:

0.19

Step-by-step explanation:

We are given that

Number of rolls of die=n=1000

Let the event of six coming up be success.Then, in each trial , the probability of success =p=P(success)=P(6)= $\frac{1}{6}$

Let X be the random variable for the number of sixes in the 1000 rolls of die.

Then, $X\sim Binom(1000,\frac{1}{6})$

Since, n is very large,the binomial random variable can be approximated as normal random variable.

Mean, $\mu=np=1000\times \frac{1}{6}=166.67$

Variance= $\sigma^2=np(1-p)=1000\times \farc{1}{6}\times (1-\frac{1}{6})=1000\times \frac{5}{36}=138.89$

$X\sim N(166.67,138.89)$

$P(150\leq X\leq 200)=P[\frac{150-166.67}{11.79}\leq \frac{X-\mu}{\sigma}\leq \frac{200-166.67}{11.79}]$

= $P[-1.41\leq Z\leq 2.83]=P[Z\leq 2.83]-P[Z$

= $\phi(2.83)-\phi(-1.41)$

=0.9977-0.0793=0.9184

Thus, the probability that the number 6 appears between 150 to 200 times=0.92

Now, given that 6 appears exactly 200 times .

Therefore, other number appear in other 800 rolls .

We have to find the probability that the number 5 will appear less than 150 times.

Therefore, for 800 rolls, let the event of 5 coming up be success.

Then , p=P(success)=P(5)= $\frac{1}{5}$

Let Y be the random variable denoting the number of times 5 coming up in 800 rolls.

Then, $Y\sim bin(800,\frac{1}{5})$

Mean, $\mu=np=800\times \frac{1}{5}=160$

Variance, $\sigma^2=np(1-p)=800\times \frac{1}{5}(1-\frac{1}{5})=128$

$Y\sim N(160,128)$ because n is large

$P(Y$

$P(Y$

Hence, the probability that the number 5 will appear less than 150 times given that 6 appeared exactly 200 times=0.19

4 0

2 years ago