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Dmitry [639]
3 years ago
11

1. The county fair charges $1.25 per Sicket for the rides. Jermaine bought 25 tickets for the rides and spent a

Mathematics
1 answer:
schepotkina [342]3 years ago
6 0

Answer:

A

Step-by-step explanation:

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Answer:

x/9 = 1/5

x = 9/5

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Felipe Rivera's savings account has a balance of 2163 After 3 years what will the amount of interest be at 4%
Allisa [31]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad  \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2163\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years\dotfill &3 \end{cases} \\\\\\ A=2163\left(1+\frac{0.04}{4}\right)^{4\cdot 3}\implies A=2163(1.01)^{12}\implies A\approx 2437.322540175

3 0
3 years ago
Convert the following statement using an "if-then" structure.
GaryK [48]

Answer:

  D. If John owns a dog, then he owns a cat

Step-by-step explanation:

The implication p → q (if p, then q) has the same truth table as the logical expression ~p∨q. You have the expression ...

  ~(John owns a dog) ∨ (he owns a cat)

Matching parts of this expression to the components of the expression ~p∨q, we see we can choose ...

  • p = John owns a dog
  • q = he owns a cat

and put those into the structure of the implication: if p, then q.

  If John owns a dog, then he owns a cat. . . . . matches choice D

6 0
3 years ago
Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea
elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) =  P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

  • The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
  • There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways to select exactly 3 aces is:

4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052

5 0
3 years ago
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