All categories

3 years ago

8k-6=(k+3)+6k solving equations with distributive property.

Mathematics

1 answer:

julsineya [31]3 years ago

5 0

Answer:

k=9

Step-by-step explanation:

8k-6=(k+3)+6k first remove the unnecsarry parentheses which would make it 8k-6=k+3+6k

then collect the like terms and get 8k-6=7k+3 then move the variable to the left and change its sign to get 8k-7k-6=3.

then collect the like terms a final time which would get you k=3+6 to then get k=9

hope this helps!

You might be interested in

Of 20 people who drink coffe 12 don’t put sugar what fraction would represent the people that use sugar in there coffee

Mama L [17]

Answer:

Fraction 8/20

Step-by-step explanation:

So only 12 people don't use sugar so if we were to put a fraction for people who don't put sugar then it would be.

12/20

The 20 stands about how many people their are and the 12 stands for how many people don't put sugar.

So now we are supposed ot do peope who put sugar so it would look like this

8/20

How I got the 8 is that I took the 20 people and subtrack it with the people who don't get sugar which is 8 and 8 people use sugar so theirs our answers.

8/20

8 0

2 years ago

The sample space, S, of a coin being tossed three times is shown below, where H and T denote the coin landing on heads and tails

Lera25 [3.4K]

Answer:

The probability distribution for the number of heads occurring in three coin tosses are P(X1) = 1, P(X2)= 2/3, P(X3)= 2/3, P(X4)= 1/3, P(X5)= 2/3, P(X6)= 1/3, P(X7)= 1/3 and P(X8)= 0

Step-by-step explanation:

The question is incomplete. Here is the complete question;

The sample space, S, of a coin being tossed three times is shown below, where Hand T denote the coin

landing on heads and tails respectively.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Let X = the number of times the coin comes up heads. What is the probability distribution for the number of heads occurring in three coin tosses?

Probability is the likelihood or chance that an event will occur.

Probability = expected outcome/total outcome of event

From the question given, the total number of sample space will be our total outcome during each toss = 3

If X is the number of times the coin comes up heads, the number of times the coin turn up all heads will be the expected outcome during each throw.

Probability distribution P(Xi)= $\frac{n(Xi)}{n(S)}$

Xi is the probability that the head turns up at each throw.

n(Xi) is the expected outcome during each toss

n(S) is the total outcome during each toss

During the first throw, the head turns up three times (HHH), the probability that heads turns up ia 3/3 = 1

P(X1) = 1

During the second, third and fifth throw, the head turns up two times, the probability that heads turns up during this times will be 2/3

P(X2)= 2/3

P(X3)= 2/3

P(X5)= 2/3

During the fourth, sixth and seventh throw, the head turns up just once, the probability that heads turns up during this times will be 1/3

P(X4)= 1/3

P(X6)= 1/3

P(X7)= 1/3

Since no head turns up during the eight throw, the probability that heads turns up during this time will be 0/3 which is 0.

P(X8)= 0

The probability distribution for the number of heads occurring in three coin tosses are P(X1) = 1, P(X2)= 2/3, P(X3)= 2/3, P(X4)= 1/3, P(X5)= 2/3, P(X6)= 1/3, P(X7)= 1/3 and P(X8)= 0