Juhi G.

Graduate Assistant In Mathematics department, Jan-15 - Present, Illinois State University

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Statistics

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Question:

A card is drawn at random from a deck of cards. Find the probability of getting a queen.

Juhi G.

Answer:

Step-1: There are total 52 cards in a deck and 4 cards are of queen from those 52 cards Step-2: We are asked to find the probability of getting a queen out of those 52 cards Step-3: In our case, the sample space would be the whole deck of cards, so the number of cards/number of the elements in our sample space would be n(S)=52 Step-4: Our event is to get a queen, and there are 4 queens in a deck of 52 cards so the number of elements in the event set would be n(E)=4 Step-5: probability of any event is P(E)=n(E)/n(S) Step-6: So it is equal to 4/52=1/13 Step-7: So, our answer is 1/13

Algebra

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Question:

Given: h(t) = t + 2 + 3; Find h(6)

Juhi G.

Answer:

Step-1: We are given a function in t and we are supposed to find the value of that function when its independent variable that is t becomes equal to 6 Step-2: So, lets plug in the value t=6 in the expression h(t) = t + 2 + 3 Step-3: So, h(6)=6+2+3 Step-4: Adding all numbers, We get h(t)= 11

Algebra

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Question:

Evaluate 5q^2+2q-3=0?

Juhi G.

Answer:

Step-1: This is a quadratic equation, because the highest power of q is 2, so we will be getting two solutions for q Step-2: We need to apply the AC method, so the product of the roots should be (5*(-3))=-15 and the sum/difference of the roots should be (5-3)=2 Step-3: If we split the middle term 2q as 5q-3q, we can have the product as -15 and sum as 2 Step-4: Rewriting the equation as: 5q^2+5q-3q-3=0 Step-5: Taking 5q common from first two terms and -3 common from last two terms, we can write our equation as 5q(q+1)-3(q+1)=0 Step-6: Taking (q+1) common, we are left with (5q-3)(q+1)=0 Step-7: Equating both the parenthesis to 0 we get two solutions for q as q=3/5 and q=-1 Step-8: Verification for the ans: If we plug in back the answers to the equation, we will get left hand side term equal to right hand side term, thus that proves our solution is right.

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