1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
navik [9.2K]
2 years ago
13

8+(x divided by 3)=-2

Mathematics
2 answers:
KIM [24]2 years ago
8 0

Answer:

x=-6

Step-by-step explanation:

Given: 8+ (x/3)=-2

Step 1: 8+(x/3 times 3) = -2 times 3 which leads to 8+x=-6

Step 2: 8(-8)+x=-6 which leads to x=-6

Answer: x=-6

fenix001 [56]2 years ago
7 0

Answer:

x = -18

Step-by-step explanation:

8 + x/3=-2

3×8+x/3×3=-2×3

24+x=-6

x=-18

You might be interested in
Answer the questions below based on a bag of Jolly Ranchers having the following number of each flavor:
viva [34]
Part A) About 19/100
Part B) About 4/25
Part C) About 7/20
Part D) About 13/20
8 0
2 years ago
Solve Linear Equations with Rational Numbers
german

Answer:

3/2

1x = 12

Step-by-step explanation:

See attachment

5 0
2 years ago
How do you know that the bowling pins are set up in parallel lines?
polet [3.4K]

Answer:

you look at them from an angle

5 0
2 years ago
A 52-card deck is thoroughly shuffled and you are dealt a hand of 13 cards. (a) If you have at least one ace, what is the probab
jasenka [17]

Answer:

a) 0.371

b) 0.561

Step-by-step explanation:

We can answer both questions using conditional probability.

(a) We need to calculate the probability of obtaining two aces given that you obtained at least one. Let's call <em>A</em> the random variable that determines how many Aces you have. A is a discrete variable that can take any integer value from 0 to 4. We need to calculate

P(A \geq 2 | A \geq 1) = P(A\geq 2 \cap A \geq 1) / P(A \geq 1)

Since having 2 or more aces implies having at least one, the event A \geq 2 \cap A \geq 1 is equal to the event A \geq 2. Therefore, we can rewrite the previous expression as follows

P(A \geq 2) / P(A \geq 1)

We can calculate each of the probabilities by substracting from one the probability of its complementary event, which  are easier to compute

P(A \geq 2) = 1 - P((A \geq 2)^c) = 1 - P((A = 0) \bigsqcup (A = 1)) = 1 - P(A = 0) - P (A = 1)

P (A \geq 1) = 1 - P ((A \geq 1)^c) = 1 - P(A = 0)

We have now to calculate P(A = 0) and P(A = 1).

For the event A = 0, we have to pick 13 cards and obtain no ace at all. Since there are 4 aces on the deck, we need to pick 13 cards from a specific group of 48. The total of favourable cases is equivalent to the ammount of subsets of 13 elements of a set of 48, in other words it is 48 \choose 13. The total of cases is 52 \choose 13. We obtain

P(A = 0) = {48 \choose 13}/{52 \choose 13} = \frac{48! * 39!}{52!*35!} \simeq 0.303  

For the event A = 1, we pick an Ace first, then we pick 12 cards that are no aces. Since we can pick from 4 aces, that would multiply the favourable cases by 4, so we conclude

P(A=1) = 4*{48 \choose 12}/{52 \choose 13} = \frac{4*13*48! * 39!}{52!*36!} \simeq 0.438      

Hence,  

1 - P(A = 1)-P(A=0) /1-P(A=1) = 1 - 0.438 - 0.303/1-0.303 = 0.371

We conclude that the probability of having two aces provided we have one is 0.371

b) For this problem, since we are guaranteed to obtain the ace of spades, we can concentrate on the other 12 cards instead. Those 12 cards have to contain at least one ace (other that the ace of spades).

We can interpret this problem as if we would have removed the ace of spades from the deck and we are dealt 12 cards instead of 13. We need at least one of the 3 remaining aces. We will use the random variable B defined by the amount of aces we have other that the ace of spades. We have to calculate the probability of B being greater or equal than 1. In order to calculate that we can compute the probability of the <em>complementary set</em> and substract that number from 1.

P(B \geq 1) = 1-P(B=0)

In order to calculate P(B=0), we consider the number of favourable cases in which we dont have aces. That number is equal to the amount of subsets of 12 elements from a set with 48 (the deck without aces). Then, the amount of favourable cases is 48 \choose 12. Without the ace of spades, we have 51 cards on the deck, therefore

P(B = 0) = {48 \choose 12} / {51 \choose 12} = \frac{48!*39!}{51!*36!} = 0.438

We can conclude

P(B \geq 1) = 1- 0.438 = 0.561

The probability to obtain at least 2 aces if we have the ace of spades is 0.561

4 0
3 years ago
Which is equivalent to (3 + 2i)(5 + i)?
zepelin [54]
(3 + 2i)(5 + i)
15 + 3i + 10i + 2i^2
15 + 13i - 2
13 + 13i
5 0
2 years ago
Other questions:
  • If Root (3/5)x+1 = 125/27 find the value of x.
    11·1 answer
  • A certain pet store has only cats and dogs. The ratio of the number of cats to the
    10·2 answers
  • Find the volume of the prism.CmCm
    7·1 answer
  • How shall i find the square root of 169 without using calculator?? please describe briefly..
    12·1 answer
  • What is the value of x that will make A parallel to B?
    12·1 answer
  • Help plz I dont understand ​
    9·2 answers
  • Integrate with respect to x <br><br> 3√(2-5x)
    9·1 answer
  • Which of the following are right triangle congruence theorems
    13·1 answer
  • On Saturday morning, Owen earned $24 raking leaves. By the end of the afternoon he had earned a total of $62. Write an equation
    15·1 answer
  • An egg-packaging factory can package eggs in cartons that hold 12 eggs or in cartons that hold 18 eggs.
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!