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nirvana33 [79]
3 years ago
5

Ill give brainliest please help with this algebra!!

Mathematics
2 answers:
aev [14]3 years ago
4 0
So u times the 8 by 2 which gives u 16 then u simplify it then u get 2/4
enot [183]3 years ago
3 0

Answer:

You didnt post a picture of the 3 expressions so therefor I cant help but if you post them then I can try to help. :)

Step-by-step explanation:

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In a sale a gameboys are reduced by two fifths. what is the sale price if the original price was 50
vladimir1956 [14]

Answer:

The sale price would be $30.

Step-by-step explanation:

2/5 is like saying 20/50. If you take 50 and subtract 20 from that, you will be left with 30.

4 0
3 years ago
If you make a tree diagram and then list all of the possible outcomes for flipping a coin and rolling a 6-sided die, how many po
Katyanochek1 [597]

Answer:

The possible outcomes that will be obtained are:

Option: C

C) 12

Step-by-step explanation:

When we flip a coin the outcomes that are obtained are:

{H,T}

Now, the possible outcomes that is obtained on flipping a die after flipping a coin are:

{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

Hence, the total number of possible outcomes that are obtained on flipping a coin and rolling a six-sided die are:

12

Also the tree diagram of these possible outcomes is attached to the answer.

4 0
3 years ago
Is 2.11 a repeating decimal
Sergeu [11.5K]

Answer:

No, 2.11 is not a repeating decimal. A repeating decimal is one which goes on forever, but you can easily see that 2.11 ends.

Hope this helps :)

4 0
4 years ago
Read 2 more answers
Urgent!!! please help!!
serious [3.7K]

Answer:

can you please tell me what grade you're in like

what is

7 0
3 years ago
You are at a stall at a fair where you have to throw a ball at a target. There are two versions of the game. In the first
Tomtit [17]

Answer:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

Alternative 1

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

X \sim Binom(n=3, p=0.1)

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!}

We can find the probability of loss like this P(X=0) and if we find this probability we got this:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

Alternative 2

Let Y the random variable of interest, on this case we now that:

Y \sim Binom(n=5, p=0.05)

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

P(Y)=(nCy)(p)^y (1-p)^{n-y}

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

nCy=\frac{n!}{(n-y)! y!}

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

4 0
4 years ago
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