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
NikAS [45]
3 years ago
9

Which of the following expressions represents the phrase “twice the sum of a number and 5”

Mathematics
1 answer:
olga nikolaevna [1]3 years ago
5 0

Answer:

2(x+5)

Step-by-step explanation:

You might be interested in
Need help solving this please
Aloiza [94]

Answer:

A) x=39

Step-by-step explanation:

117 = 3x

117 divided by 3 = 39

3x = 117

3(39) = 117

117 = 117

A) x = 39

4 0
2 years ago
Find the slope. ASAP
vagabundo [1.1K]
The answer is number 4
3 0
3 years ago
Which of the following is a type of data that is likely to be normally<br> distributed?
kkurt [141]
It is either A or D, but I think it’s A
8 0
2 years ago
Read 2 more answers
(x+2)*2=2, x=? I need help
Agata [3.3K]

Answer:

2x2−x2 2 x 2 - x 2. Subtract x2 x 2 from 2x2 2 x 2 . x2 x 2. 2x2−x2 2 x 2 - x 2. (. [. ([. ) ] )] |. |. √. √... > ≥. >≥.......... 7. 7. 8. 8. 9. 9.

Step-by-step explanation:

work it out dude

5 0
3 years ago
Read 2 more answers
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
Other questions:
  • Zach walks dogs for $20 per month per dog. He also has a part-time job that pays $9.70 per hour. Zach’s car payment and insuranc
    13·1 answer
  • PLZ HELP ITS DUE TOMORROW ​
    10·1 answer
  • often there are multiple ways to prove that lines are parallel. how easily do you recognize different possibilities
    13·2 answers
  • The average house in a neighborhood measures 55 ft by 35 ft by 20 ft tall. What is the total outside surface area?
    5·1 answer
  • Which numbers are greater than 3x10^-7
    13·1 answer
  • 5 divided by the sum of a and b
    7·2 answers
  • Mr. Singleton wants to buy a new television. The regular price is $999. The store is offering a 25% discount and a sales tax of
    14·1 answer
  • Please answer this the best you possible can :)
    13·1 answer
  • Help?<br> ------------------
    10·2 answers
  • A bag holds 13 marbles. 6 are blue, 2 are green, and 5 are red. If you select two marbles from the bag in a row without replacin
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!