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
shtirl [24]
3 years ago
9

55% of 19.6 is what number?

Mathematics
1 answer:
tiny-mole [99]3 years ago
3 0
10.78 is the answer.
19.6 x .55
You might be interested in
Consider the pattern shown in the table below
irina [24]

Answer:

there is not table

Step-by-step explanation:

4 0
3 years ago
Find the value of x.
jok3333 [9.3K]

Answer:

Complementary

Step-by-step explanation:

5 0
2 years ago
A person is 6 feet tall and casts a shadow of 3.2 ft at the same time a tree casts a shadow of 36.8 ft how tall is the tree?
Elan Coil [88]

Answer:

18.5

Step-by-step explanation:

5 0
3 years ago
In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
Marrrta [24]

Answer:

a) Bi [P ( X >=15 ) ] ≈ 0.9944

b) Bi [P ( X >=30 ) ] ≈ 0.3182

c)  Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) Bi [P ( X >40 ) ] ≈ 0.0046  

Step-by-step explanation:

Given:

- Total sample size n = 745

- The probability of success p = 0.037

- The probability of failure q = 0.963

Find:

a. 15 or more will live beyond their 90th birthday

b. 30 or more will live beyond their 90th birthday

c. between 25 and 35 will live beyond their 90th birthday

d. more than 40 will live beyond their 90th birthday

Solution:

- The condition for normal approximation to binomial distribution:                                                

                    n*p = 745*0.037 = 27.565 > 5

                    n*q = 745*0.963 = 717.435 > 5

                    Normal Approximation is valid.

a) P ( X >= 15 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]

 - Then the parameters u mean and σ standard deviation for normal distribution are:

                u = n*p = 27.565

                σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522

- The random variable has approximated normal distribution as follows:

                X~N ( 27.565 , 5.1522^2 )

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= -2.5358 ) = 0.9944

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944

Hence,

                Bi [P ( X >=15 ) ] ≈ 0.9944

b) P ( X >= 30 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= 0.37556 ) = 0.3182

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182

Hence,

                Bi [P ( X >=30 ) ] ≈ 0.3182  

c) P ( 25=< X =< 35 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )

                N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

- Now use the Z-score table to evaluate the probability:

                P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

               N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

Hence,

                Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) P ( X > 40 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )

                N [ P ( X > 41 ) ] = P ( Z > 2.60762 )

- Now use the Z-score table to evaluate the probability:

               P ( Z > 2.60762 ) = 0.0046

               N [ P ( X > 41 ) ] =  P ( Z > 2.60762 ) = 0.0046

Hence,

                Bi [P ( X >40 ) ] ≈ 0.0046  

4 0
3 years ago
What is the total cost of 0.5 pound of peaches selling for $0.80 per pound and 0.7 pound of oranges selling for $0.90 per pound?
swat32
Given situation : 0.5 pound of peaches selling for 0.80 dollars/ pound 0.7 pound of oranges selling for 0.90 dollars / pound. Solution Given number 1 : Peaches => 0.5 pound = meaning, ½ pound is available, And 1 pound of it costs 0.80 dollars. Let’s solve: => 0.80 dollars * 0.5 => 0.40 dollars – the price of the peaches. Given number 2 : Oranges => 0.7 pounds of oranges, meaning less than 1 pound. And 1 pound costs 0.90 dollars Let’s solve to get the anwer => 0.7 * .90 => 0.63 dollars – the costs of 0.7 pounds of oranges,
7 0
3 years ago
Other questions:
  • What adds up to 14 and multiplies to negative 576?
    8·2 answers
  • Tara has a large box of dog treats that weighs 8.4 pounds. She uses the large box of dog treats to make smaller bags, each conta
    11·2 answers
  • Write £1.44 to the nearest 10pence
    7·1 answer
  • How much money do you earn in 1 hour if you earn 20 in 4 hours
    15·2 answers
  • Elizabeth brought a box of donuts to share. There are​ two-dozen (24) donuts in the​ box, all identical in​ size, shape, and col
    10·1 answer
  • If three candys cost four dollars how much would two candy cost?
    10·2 answers
  • If 7x+2a=3x+5a then x is equivalent to what
    7·2 answers
  • PLEASE HELP! DUE TODAY :))
    11·1 answer
  • In football, the team that has the ball has four chances to gain at least ten yards, if
    5·1 answer
  • Can someone help me with this problem? Thank you!
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!