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Sunny_sXe [5.5K]
3 years ago
5

Create an expression that when combined equals 5x+2y (must be at least 3 terms)

Mathematics
1 answer:
Radda [10]3 years ago
8 0
You could do 3x+y+2x+y
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Please help me please
olya-2409 [2.1K]

Answer:

its 17

Step-by-step explanation:

do 2 step Equation

3 0
2 years ago
Heather saves $5.00 on Monday. Each day after that, she decides to save twice as much as she did the previous day. If she saves
Troyanec [42]
5+10+20+40=75..........
5 0
3 years ago
Read 2 more answers
Which of the following numbers is between 3/5 and 5/7
Vsevolod [243]

Numbers 8/9 and 47/70 are between 3/5 and 5/7

Step-by-step explanation:

  • Step 1: Find the equivalent decimal expressions of all the given fractions to find which ones are between 3/5 and 5/7

3/5 = 0.6 and 5/7 = 0.71

Option 1 ⇒ 1/2 = 0.5

Option 2 ⇒ 3/7 = 0.43

Option 3 ⇒ 8/9 = 0.89

Option 4 ⇒ 19/35 = 0.54

Option 5 ⇒ 47/70 = 0.67

∴ 8/9 and 47/70 lies between 3/5 and 5/7

5 0
3 years ago
What does x equal in this equation
nataly862011 [7]
What’s the equation? i don’t see it?
4 0
3 years ago
Read 2 more answers
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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