1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Kitty [74]
3 years ago
10

When a fisherman catches a fish, the probability that it is a young one is 0.23 . All young fishes are returned to the water. On

the other hand, an adult fish is kept for eating later. Out of the five fishes caught by the fisherman, What is the probability that at most 3 are edible
Mathematics
1 answer:
max2010maxim [7]3 years ago
4 0

Answer:

The value is  P(X = 3) =   0.0725  

Step-by-step explanation:

From the question we are told that

   The probability catching a young fish is   q =  0.23

    The number of fish that was caught is  n = 5  

Generally the probability of catching a adult fish is mathematically represented as

     p =  1 -q

=>   p =  1 - 0.23      

=>   p =  0.77

Generally for a fish to be edible it must be an adult fish.

Generally the distribution of the number of fish the fisher man caught  follows a binomial distribution  

i.e  

         X  \~ \ \ \  B(n , p)

and the probability distribution function for binomial  distribution is  

      P(X = x) =  ^{n}C_x *  p^x *  (1- p)^{n-x}

Here C stands for combination hence we are going to be making use of the combination function in our calculators  

So

        P(X = 3) = \  ^5C_3  *(0.77)^{3} *  (0.23)^{5- 3 }  

=>     P(X = 3) =   3  * 0.45653 *  0.0529  

=>     P(X = 3) =   0.0725  

You might be interested in
2+363892917382827363
pentagon [3]

Answer:

363892917382827365

Step-by-step explanation:

thats how one adds

3 0
3 years ago
Read 2 more answers
X10.IN_4.PS-11
Basile [38]

Answer:

sorry but I really don't know the answer.

Step-by-step explanation:

8 0
3 years ago
Please help thank you so much
ahrayia [7]
So meagure the b side than the a side than theres ur answer
5 0
3 years ago
What are the next 3 numbers in the sequence... 5,25,125...
OLga [1]

5 . . . . . is multiplied by 5 to get . . . . . 25

25 . . . . . is multiplied by 5 to get . . . . . 125

so

125 . . . . . is multiplied by 5 to get . . . . . 625

625 . . . . . is multiplied by 5 to get . . . . . 3125

and

3125 . . . . . is multiplied by 5 to get . . . . . 15,625
8 0
3 years ago
starting at home, nadia traveled uphill to the grocery store for 30 minutes at just 4 mph. she then traveled back home along the
ankoles [38]
Distance from home to store = 4 * 0.5 = 2 miles.
Time taken to return home = (2/12) * 60 =10 minutes.
Total distance traveled = 2 * 2 = 4 miles.
Total time taken = 30 + 10 = 40 minutes.
Average speed for entire trip = total distance / total time in hours.
Average\ speed=\frac{4}{\frac{40}{60}}=\frac{4\times60}{60}=6\ mph
3 0
4 years ago
Read 2 more answers
Other questions:
  • I need help with this math problem
    14·1 answer
  • Definition for compatible numbers
    13·1 answer
  • Can you please simplify: sqrt 75x^6
    8·1 answer
  • The function f(x) = 3x^2 models the packaging costs, in cents, for a box shaped like a rectangular prism. The side lengths are 4
    13·1 answer
  • A long wire is wound in a circular coil of a radius of 1 meter. The coil consist of 70 loops of the wire are around it. What is
    11·1 answer
  • Write the expression as either the sine, cosine, or tangent of a single angle.
    7·2 answers
  • 3 Times Square root of 200 simplified
    8·2 answers
  • A football team loses 3 yards, gains 12 yards, gains 10 yards, and then loses 15 yards. Did the team gain yards or lose yards ov
    11·1 answer
  • In an old photograph, the height of grandma's table is 3.8 cm. The table doesn't exist anymore; grandma doesn't know what happen
    8·1 answer
  • segment MN is shown. Point M is located at (7,6). Point N is located at (7,-4) What is the midpoint segment of MN
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!