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Kruka [31]
3 years ago
8

time it takes to fly from City A to City B a. Is the distance a baseball travels in the air after being hit a discrete random​ v

ariable, a continuous random​ variable, or not a random​ variable?
Mathematics
1 answer:
Tresset [83]3 years ago
7 0

Complete question is;

Determine whether the random variable is discrete or continuous.

a. The time it takes to fly from City Upper A to City Upper B.

b. The time it takes for a light bulb to burn out.

c. The number of fish caught during a fishing tournament.

d. The number of textbook authors now sitting at a computer.

e. The distance a baseball travels in the air after being hit.

a. Is the time it takes to fly from City Upper A to City Upper B discrete or​ continuous? A. The random variable is discrete. B. The random variable is continuous.

Answer:

a) continuous random variable.

b) continuous random variable

c) discrete random variable

d) discrete random variable

e) continuous random variable

Step-by-step explanation:

A discrete variable is one where the value of the variable is gotten by counting the variables.

Meanwhile, a continuous variable is one where the value of the variable is gotten by measuring the variables.

For options a & b, the time taken is the variable that is measured. Thus, they are both continuous random variable.

Also, for option e, the distance traveled is the measured variable and thus it is also a continuous random variable.

In option c, the number of fishes can be counted. Also, for option d, the number of textbook authors can be counted. Thus, the values of the variables can be gotten by counting. . Therefore, they are both discrete random variable.

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Is 3/15 repeating decimal
sp2606 [1]

Answer:

no

Step-by-step explanation:

3/15 is .2

8 0
3 years ago
A school has 8 blocks all day that last for 45 minutes each day. How long would each block be if school had 9 blocks all day
katrin2010 [14]

Answer: (reduce if needed)

50.625

Step-by-step explanation:

45 divided by 8 is 5.625

Add 5.625 to 45 for 50.625

Mark me as brainliest if this helps!

3 0
3 years ago
Let X be a binomial random variable with p = 0.7 and n = 10. Calculate the following probabilities from the binomial probability
padilas [110]

Answer:

0.4114  

0.0006  

0.1091  

0.1957  

Step-by-step explanation:

<u>Given:  </u>

p = 0.7 n = 10

We need to determine the probabilities using table , which contains the CUMULATIVE probabilities P(X \leq x).  

a. The probability is given in the row with n = 10 (subsection x = 3) and in the column with p = 0.7 of table:  

P(X \leq  3) = 0.4114  

b. Complement rule:  

P( not A) = 1 - P(A)

Determine the probability given in the row with n = 10 (subsection x = 10) and in the column with p = 0.7 of table:  

P(X \leq  10) = 0.9994

Use the complement rule to determine the probability:  

P(X > 10) = 1 - P(X\leq 10) = 1 - 0.9994 = 0.0006  

c. Determine the probability given in the row with n = 10 (subsection x = 5 and x = 6) and in the column with p = 0.7 of table:  

P(X \leq  5) = 0.8042

P(X \leq  6) = 0.9133

The probability at X = 6 is then the difference of the cumulative probabilities:  

P(X = 6) = P(X \leq  6) - P(X \leq  5) = 0.9133 — 0.8042 = 0.1091  

d. Determine the probability given in the row with n = 10 (subsection x = 5 and x = 11) and in the column with p = 0.7 of table:  

P(X \leq  5) = 0.8042

P(X \leq  11) = 0.9999

The probability at 6 \leq X \leq 11 is then the difference between the corresponding cumulative probabilities:  

P(6 \leq  X \leq 11) = P(X \leq 11) - P(X \leq  5) = 0.9999 — 0.8042 = 0.1957  

6 0
3 years ago
Please help me with this
tester [92]

Answer:

(3,1) and (0,2)

Step-by-step explanation:

6 0
3 years ago
A manufacturer makes three types of screws. Type A comes in a bulk pack of 1000 screws. Type B comes in a bulk pack of 500 screw
Tpy6a [65]

Answer:

Step-by-step explanation:

Given that a manufacturer makes three types of screws

Type             A        B           C Total

   

Pack         1000 500   800  

P for flaw 0.001 0.03 0.005  

Pack*p                 1 15                    4     20

a) Expected total number of defective screws = 20

b) For one pack each no of defective screws =20

Hence for 5 expected number number of packs form each should be 1/3

8 0
3 years ago
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