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S_A_V [24]
2 years ago
15

Im wasting so many points on these help please

Mathematics
1 answer:
nata0808 [166]2 years ago
7 0

Hello, again!! :p

I'm not sure about the whole equation because I was taught to solve it in a different way and I got the answer  =3 ± i√ 6 ,not a -3.

Sorry, I'm not sure if that was much help!

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A coin is tossed 40 times and shows heads 23 times. Calculate the experimental probability of the coin showing heads.
Over [174]

Answer:

0.575

Step-by-step explanation:

There are two types of probabilities:

1. Theoretical Probability

2. Experimental Probability

The probability based on mathematical theories is called theoretical probability. It is not always true. i.e. the probability of getting a head in a toss of coin is 1/2 or 0.5 or 50%

Experimental probability is based on experiments. Experimental probability is always going to change whenever the experiment is performed.

The theoretical probability of getting a head while tossing a coin will always be 50% while experimental probability might be different the next time, experiment is performed.

In this case, the experimental probability will be 23/40 or 0.575 ..

8 0
3 years ago
11/12 divided by 4 as a fraction
otez555 [7]

Answer: 11/48

Step-by-step explanation:

(11/12) / 4 = 11/12 * 1/4 = (11*1)/(12*4) = 11/48

4 0
3 years ago
Read 2 more answers
Which sentence represents a relationship between the number of visits
Step2247 [10]
Choice (A) has zero slope becsuse her cost doesn't depend on the number of visits.
6 0
3 years ago
Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if
Delicious77 [7]

Answer:

(1)Equivalence Relation

(2)Not Transitive, (0,3) is missing

(3)Equivalence Relation

(4)Not symmetric and Not Transitive, (2,1) is not in the set

Step-by-step explanation:

A set is said to be an equivalence relation if it satisfies the following conditions:

  • Reflexivity: If \forall x \in A, x \rightarrow x
  • Symmetry: \forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x
  • Transitivity: \forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z

(1) {(0,0), (1,1), (2,2), (3,3)}

(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}

The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.

(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}

In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.

Therefore, It is not transitive.

As a result, the set (2) is not an equivalence relation.

(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}

(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric

Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.

5 0
3 years ago
What is the range of the function on the graph?
Marta_Voda [28]

Answer:

no. 4

Step-by-step explanation:

the numbers start at two and keeps going up.

3 0
3 years ago
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