PLS HELP ME ASAP!!!!!!!!!!!!!!

Five years after 650 high school seniors graduated, 400 had a college degree and 310 were married. Half of the students with a c

DedPeter [7]

Answer:

The probability that a student has a college degree or is not married is 0.8308.

Step-by-step explanation:

The information provided is:

Total number of high school seniors (N) = 650.

Number of seniors with a college degree (n (C)) = 400.

Number of seniors who were married, (n (M)) = 310.

Consider the Venn diagram below.

The probability of an event, say E, is the ratio of the favorable outcomes of E to the total number of outcomes of the experiment.

That is,

$P(E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes of E

N = total number of outcomes of the experiment.

The probability of the union of two events is:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{n(A)+n(B)-n(A\cap B)}{N}$

Compute the probability that a student has a college degree or is not married as follows:

$P(C\cup M^{c})=\frac{n(C)+n(M^{c})-n(C\cap M^{c})}{N}$

From the Venn diagram:

n (C) = 400

n ( $M^{c}$ ) = N - n (M) = 650 - 310 = 340

n (C ∩ $M^{c}$ ) = 200

The value of P (C ∪ $M^{c}$ ) is:

$P(C\cup M^{c})=\frac{n(C)+n(M^{c})-n(C\cap M^{c})}{N}=\frac{400+340-200}{650}=0.83077\approx0.8308$

Thus, the probability that a student has a college degree or is not married is 0.8308.

6 0

3 years ago

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A segment in the complex plane has a midpoint at –1 + i. If the segment has an endpoint at –5 – 7i, what is the other endpoint?

saul85 [17]

Answer:

Option D. 3 + 9i

Step-by-step explanation:

We know that the formula to find the middle point is:

$(\frac{x_1 + x_2}{2},\frac{z_1 + z_2}{2})$

We know that the middle point is:

$-1 + i$

And the end point $(x_2, z_2)$ is: $-5 -7i$

We wish to find the point $x_1,z_1$

So we can write the following equation

$\frac{x_1 + (-5)}{2} = -1$ (i)

and

$\frac{z_1 + (-7i)}{2} = i$ (ii)

Now we clear $x_1$ from equation (i) and clear $z_1$ from equation (ii)

For equation (i) we have:

$-2 = x_1 -5 \\\\x_1 = 3$

For equation (ii) we have:

$2i = z_1 -7i\\\\z_1 = 9i$

Then the initial point $x_1, z_1$ is: $3 + 9i$

The correct answer is the last option 3 +9i

3 0

3 years ago

(GIVING BRAINLIEST!!!)

Ivenika [448]

Answer:

the answer is c

Step-by-step explanation:

hope i helped

4 0

3 years ago

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Suppose a certain state university's college of business obtained the following results on the salaries of a recent graduating c

Lerok [7]

Answer:

Following are the responses to the given choice:

Step-by-step explanation:

For point a:

$H_0: \mu_1 - \mu_2 = 0\\\\ H_1: \mu_1 - \mu_2 < 0$

For point b:

$t = -2.953$

For point c:

$\to p- value = 0.0021$

For point d:

Reject $H_o$ . It could deduce that the pay of higher banking is considerably lower than the pay of higher project management.

7 0

3 years ago

James cut out four parallelograms, the dimensions of which are shown below. Parallelogram 1 length: 12 in. width: 15 in. diagona

MrRa [10]

Answer: The correct options are;

*The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 1 do not form right angles

**The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 4 do not form right angles.

Step-by-step explanation: What James is trying to do is quite simple which is, he wants to place four quadrilaterals inside a circle and he wants the vertices to touch one another at the center of the circle without having to overlap.

This is possible and quite simple, provided all the quadrilaterals have right angles (90 degrees). This is because the center of the circle measures 360 degrees and we can only have four vertices placed there without overlapping only if they all measure 90 degrees each (that is, 90 times 4 equals 360).

We can now show whether or not all four parallelograms have right angles by applying the Pythagoras' theorem to each of them. Note that James has cut the shapes in such a way that the hypotenuse (diagonal) and the other two legs have already been given in the question. As a reminder, the Pythagoras' theorem is given as,

AC² = AB² + BC² Where AC is the hypotenuse (diagonal) and AB and BC are the other two legs. The experiment would now be as follows;

Quadrilateral 1;

20² = 12² + 15²

400 = 144 + 225

400 ≠ 369

Therefore the vertices of parallelogram 1 do not form a right angle

Quadrilateral 2;

34² = 16² + 30²

1156 = 256 + 900

1156 = 1156

Therefore the vertices of parallelogram 2 forms a right angle

Quadrilateral 3;

29² = 20² + 21²

841 = 400 + 441

841 = 841

Therefore the vertices of parallelogram 3 forms a right angle

Quadrilateral 4;

26² = 18² + 20²

676 = 324 + 400

676 ≠ 724

Therefore the vertices of parallelogram 4 do not form a right angle

The results above shows that only two of the parallelograms cut out have right angles (like a proper square or rectangle for instance), while the other two do not have right angles.

Therefore, the correct option are as follows;

The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 1 do not form right angles.

The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 4 do not form right angles.

8 0

3 years ago

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