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melisa1 [442]
2 years ago
15

Anwar selects a playing card at random from the following 6 cards: {9 of hearts, 5 of spades, 6 of hearts, 2 of spades, 4 of hea

rts, 7 of hearts} Let A be the event that Anwar selects an even-numbered card and B be the event that she chooses a heart. Which of the following statements are true?
Mathematics
1 answer:
ipn [44]2 years ago
6 0

Answer:

choose all except " The outcome of events A and B are dependent on each other"

Step-by-step explanation:

I got it right on K.A! (:

You might be interested in
The probability of the intersection of two events is known as a ________ probability.
Darya [45]
The probability of the intersection of two events is known as a joint probability.
5 0
1 year ago
Ankur estimated the quotient of 15 1/3 divided by -4 2/3 to be 3. Which best describes his error?
hichkok12 [17]

Solution:

As the two numbers 15 \frac{1}{3} {\text{and} -4\frac{2}{3}<em> </em>are mixed fractions.

15 \frac{1}{3}= \frac{46}{3} \\\\ -4\frac{2}{3}= \frac{-14}{3}

Now,  \frac{ \frac{46}{3}}{\frac{-14}{3}} = - 3

Ankur's answer= 3

The mistake he has committed , while dividing two fractions , he must have forgot that one of the fraction which is in the denominator bears negative sign  before it.

So, \frac {+}{-}= -.

Ankur forgot to put negative sign before 3.

So, The correct option which describes ankur's error is : Ankur found that the quotient of a positive number and a negative number is positive.

Option 3 is right choice.



5 0
2 years ago
Read 2 more answers
A television is 28.5 inches wide and 16 inches long.
Umnica [9.8K]

Answer:

  33 in

Step-by-step explanation:

The Pythagorean theorem tells you ...

  diagonal² = length² +width²

  diagonal² = (16 in)² +(28.5 in)² = 1068.25 in²

  diagonal = √(1068.25 in²) ≈ 32.684 in

The diagonal of the television is about 33 inches.

7 0
2 years ago
The output of a chemical process is continually monitored to ensure that the concentration remains within acceptable limits. Whe
marissa [1.9K]

Answer:

a)  0.31 = 31%

b) 0.03 = 3%

c) 0.36 = 36%

d) 2 times

Step-by-step explanation:

If F_X(x) is the cumulative distribution function of the random variable X, then by definition the probability P of the random variable is given by

P(X \leq x) = F_X(x)

If additionally the random variable is discrete (only has non-negative integers as outcomes as is the case in this problem) then

P(X=a)=F_X(a)-\lim_{x \to a^-}F_X(x)

a)

We are looking for P(X<2)

P(X < 2) = P(X\leq 2)-P(X=2)=F_X(2)-\lim_{x \to 2^-}F_X(x)=0.84-0.53=0.31

b)

In this case we want P(X>3)

P(X >3) = 1-P(X\leq 3)=1-F_X(3)=1-0.97=0.03

c)

Now, we are interested in P(X=1)

P(X =1) =F_X(1)-\lim_{x \to 1^-}F_X(x)=0.53-0.17=0.36

d)

The expected number of times that the process is recalibrated during the week is the expected value of the probability distribution:

P(X=1)+2P(X=2)+...+nP(X=n)+...

But it is easy to see that P(X=n) = 0 if n is an integer >4

So, the expected value is

P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)

We already have P(X=1) and P(X=2). Let's compute the rest

P(X =3) =F_X(3)-\lim_{x \to 3^-}F_X(x)=0.97-0.84=0.13

P(X =4) =F_X(4)-\lim_{x \to 4^-}F_X(x)=1-0.97=0.03

and the expected value is

0.36 + 2*0.53+3*0.13+4*0.03= 1.93 = 2 times rounding to the nearest integer.

8 0
2 years ago
(b) In a continuous series the average weight of some students is 75 kg and the sum of the weights is 6000 kg. Find the number o
Stolb23 [73]

Answer:

the number of students is 80

8 0
1 year ago
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