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telo118 [61]
1 year ago
11

Consider the following discrete probability distribution. x 15 22 34 40 P(X = x) 0.14 0.40 0.26 0.20 a. Is this a valid probabil

ity distribution? Yes, because the probabilities add up to 1. No, because the gaps between x values vary. b. What is the probability that the random variable X is less than 40?
Mathematics
1 answer:
saul85 [17]1 year ago
8 0

Answer:

(a) Yes, because the probabilities add up to 1.

(b) The probability that <em>X</em> < 40 is 0.80.

Step-by-step explanation:

The probability distribution of the random variable <em>X</em> is:

<em>    x</em>:  15   |   22  |   34   |   40

f (x): 0.14 | 0.40 | 0.26 | 0.20

(a)

The properties of a probability distribution are:

  1. 0 ≤ f (x) ≤ 1
  2. ∑ f (x) = 1

All the probability value are more than 0 and less than 1.

Compute the sum of all the probabilities as follows:

\sum f(X)=0.14+0.40+0.26+0.20=1

The sum of all probabilities is 1.

Thus, the probability distribution is valid.

(b)

Consider the probability distribution table.

Compute the probability of <em>X</em> < 30 as follows:

P (X < 40) = P (X = 15) + P (X = 22) + P (X = 34)

                =0.14+0.40+0.26\\=0.80

Thus, the probability that <em>X</em> < 40 is 0.80.

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The physician tells a woman who usually drinks 5 cups of coffee daily to cut down on her coffee consumption by 75%. If this woma
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Answer:

Step-by-step explanation:

5 cups per day.....1 cup = 8oz.....so 5 cups = (5 * 8) = 40 oz

cut down by 75%....means ur only drinking 25%

25% of 40 oz =

0.25(40) = 10 oz per day <===

6 0
2 years ago
In the trapezoid ABCD ( AB ∥ CD ) point M∈ AD , so that AM:MD=3:5. Line l ∥ AB and going trough point M intersects diagonal AC a
geniusboy [140]

Answer:

BC:BN=8:3

Step-by-step explanation:

ABCD is a trapezoid and there is a point m which belongs to AD such that AM:MD=3:5.Line "l" parallel to AB intersects the diagonal AC at p and BD at N.

Now, we know that the parallel lines divide the transversal into the segments with equal ratio, therefore, BN:NC=AM:MD

But, BC= BN+NC

Therefore, BC:BN=(BN+NC):BN

⇒BC:BN=(3+5):3

⇒BC:BN=8:3


4 0
1 year ago
Tom is 45 and pays $2042 on his mortgage each month while his total take home pay is $5950 per month. The national average for t
aniked [119]
More

First, you divide 2042/5950 which is about 0.4036, then you move the decimal 2 spaces to the right which is about 40.4, which means that 40% of Tom's income is his mortgage, higher than the average
4 0
1 year ago
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On a coordinate plane, square P Q R S is shown. Point P is at (4, 2), point Q is at (8, 5), point R is at (5, 9), and point S is
sergey [27]

Answer:

(D)The midpoint of both diagonals is (4 and one-half, 5 and one-half), the slope of RP is 7, and the slope of SQ is Negative one-sevenths.

Step-by-step explanation:

  • Point P is at (4, 2),
  • Point Q is at (8, 5),
  • Point R is at (5, 9), and
  • Point S is at (1, 6)

Midpoint of SQ =\frac{1}{2}(1+8,5+6)=(4.5,5.5)

Midpoint of PR =\frac{1}{2}(4+5,2+9)=(4.5,5.5)

Now, we have established that the midpoints (point of bisection) are at the same point.

Two lines are perpendicular if the slope of one is the negative reciprocal of the other.

In option D

  • Slope of RP =7
  • Slope of SQ  =-\dfrac17

Therefore, lines RP and SQ are perpendicular.

Option D is the correct option.

6 0
2 years ago
Compare the process of solving |x – 1| + 1 &lt; 15 to that of solving |x – 1| + 1 &gt; 15.
kicyunya [14]
In the first case we'd subtract 1 from both sides, obtaining |x-1|<14.

In the second case we'd also subtract 1 from both sides, and would obtain
 |x-1|>14.

What would the graphs look like?

In the first case, the graph would be on the x-axis with "center" at x=1.  From this center count 14 units to the right, and then place a circle around that location (which would be at x=15).  Next, count 14 units to the left of this center, and place a circle around that location (which would be -13).  Draw a line segment connecting the two circles.  Notice that all of the solutions are between -13 and +15, not including these endpoints.

In the second case, x has to be greater than 15 or less than -13.  Draw an arrow from x=1 to the left, and then draw a separate arrow from 15 to the right.  None of the values in between are solutions.

4 0
1 year ago
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