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grigory [225]
3 years ago
10

When solving a mass/particle or particle/mass conversion calculation what piece(s)

Mathematics
1 answer:
Lana71 [14]3 years ago
6 0

Answer:

i would cry if i had to do this tbh

Step-by-step explanation:

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What’s the answer to this problem?
Delicious77 [7]

Answer:

f (-4)=0

f (-1)=3

f (0)=0

Step-by-step explanation:

The answers are the y value for the given x values on the graph.

3 0
3 years ago
Read 2 more answers
Please help and explain.
Radda [10]

Explanation:

Vertex form of a quadratic function is given by y = a(x - h)² + k

where

1) 'a' determines if parabola is stretched or compressed.

   If a > 1 then graph is stretched by a factor of a.

   If 0 < a < 1, then graph is compressed by a factor of a.

2) If a > 0 then graph opens upwards with a happy face. (minimum)

3) If a < 0 then graph opens downwards with a sad face. (maximum)

4) (h, k) is the vertex point

5) The axis of symmetry is x = h

While solving for y = 1(x - 4)² + 3

Identify following's:

Vertex: (h, k) = (4, 3)

Axis of symmetry: x = 4

Max/Min: As here a > 0, Minimum (4, 3)

Stretch/compression: a = 1, the graph is stretched by a factor of 1.

Direction of opening: As a > 0, the graph opens upwards.

7 0
2 years ago
Find the volume of the cylinder.
monitta

Answer:

3.14×5(2)×6=471.24

Step-by-step explanation:

you have to take square of 5

4 0
3 years ago
The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that
salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

Part a

Let X the random variable of interest, on this case we know that:

X \sim Binom(n=20, p=0.411)

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

P(X\geq 1 ) = 1-P(X

And we can find the individual probability:

P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253

And then:

P(X\geq 1 ) = 1-P(X

And since we want the probability in the 3 classes we can assume independence and we got:

P= 0.99997^3 = 0.9992

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0
4 years ago
Which expression is equivalent to 2 (×+7)-18×+4/5
Ksivusya [100]

Answer:

14.8-16x

Step-by-step explanation:

2x+14-18x+0.8

14.8-16x

4 0
4 years ago
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