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spayn [35]
2 years ago
10

If the mean of 25, 28, a, 30 and 32 is 27, then find the value of a.

Mathematics
1 answer:
Anna11 [10]2 years ago
6 0

Answer:

a = 20

Step-by-step explanation:

The mean is calculated as

mean = \frac{sum}{count}

          = \frac{25+28+a+30+32}{5}  = \frac{115+a}{5}

Then

\frac{115+a}{5} = 27 ( multiply both sides by 5 to clear the fraction )

115 + a = 135 ( subtract 115 from both sides )

a = 20

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 Writing the equation of the weekly rates we have:
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 Answer:
 
An expressions that gives the total amount, in dollars, collected for weekly fees last year is:
 
20v + 10p
7 0
3 years ago
Select all the expressions that are solutions to 5=23x .
Ulleksa [173]
The answer would be c 5 divided by 2/3
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
3 years ago
I need help on the one about slope
pantera1 [17]
I think it is -5
If you divide -1/2 and then 4/1 you will get -5/1 or also known as -5
7 0
2 years ago
PLEASE HELP ASAP FOR 10 POINTS!! ❤️
Paha777 [63]

Answer:

6a^5

Step-by-step explanation:

3a²*2a³ = (2*3)(a² *a³)= 6a^{2+3}=6a^5

7 0
3 years ago
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