If there is a very strong correlation between two variables, then the coefficient of correlation must be

The measure of rqc = (x+12) and m PCq = (2x) set up and solve an equation to find the value for x

kotegsom [21]

1753 737272727278383828881829939393837377

3 0

3 years ago

Please help... this is super complicted

nataly862011 [7]

Answer:

(a) 412 +A -B

(b) 412 +A -B + (A-20) -(B+14)

(c) 378 +2A -2B

Step-by-step explanation:

Water flowing in adds to the amount in the reservoir. Water flowing out subtracts from the amount in the reservoir.

__

(a) 412 + A - B . . . . at the end of January

__

(b) Flowing in was (A -20), flowing out was (B +14). In February, we start with the amount at the end of January.

(412 +A -B) +(A-20) -(B +14)

__

(c) Simplifying the above, we have ...

412 -20 -14 +2A -2B = 378 +2A -2B

8 0

3 years ago

What expression simplifies to 8x-5

Delvig [45]

Answer:

It is in the simplest form already

Step-by-step explanation:

3 0

4 years ago

Plz help me i give brainliest

Anestetic [448]

Answer:

The median is 5

Step-by-step explanation:

The median is the value that is in the middle of a data set. One can find this value by lining up all of the values in the data set in ascending order (from smallest to largest). Then one will cross out answers on either side of the data set until one is left with a single value. In the event that one is left with two values, one will take the average of the two values. In other words, add the two values and then divide the result by 2.

The first step in finding the median is to list out all of the given data. Transcribe the visual data that is given in the problem into numerical data. In other words, rewrite the numbers represented by the dots as actual numbers.

1, 1, 3, 4, 4, 5, 6, 6, 6, 7, 9

Now take out numbers from each side until a single number is left in the middle.

1, 1, 3, 4, 4, 5, 6, 6, 6, 7, 9

1, 3, 4, 4, 5, 6, 6, 6, 7

3, 4, 4, 5, 6, 6, 6

4, 4, 5, 6, 6,

4, 5, 6,

5

5 0

3 years ago

Prove that if s is any sample space and u and v are events in s with u ⊆ v, then p(u) ≤ p(v).

aleksandrvk [35]

If $U\subseteq V$ , then either $U=V$ or $U$ is contained entirely in $V$ .

In the first case, we get $\mathbb P(U)=\mathbb P(V)$ .

In the second case, we have $\mathbb P(U)=\mathbb P(U\cap V)=\mathbb P(U\mid V)\cdot\mathbb P(V)$ , or $\mathbb P(U\mid V)=\dfrac{\mathbb P(U)}{\mathbb P(V)}$ . The LHS must be between 0 and 1, and this can only occur as long as $\mathbb P(U)\le\mathbb P(V)$ .

4 0

3 years ago