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
ehidna [41]
3 years ago
15

What is the sum when 1/5 is added to 5/2?

Mathematics
1 answer:
NeTakaya3 years ago
6 0

Answer:

27/10

Step-by-step explanation:

You might be interested in
Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50
Vladimir79 [104]

Answer:

a) Var[z] = 1600

    D[z] = 40

b) Var[z] = 2304

    D[z] = 48

c) Var[z] = 80

    D[z] = 8.94

d) Var[z] = 80

    D[z] = 8.94

e) Var[z] = 320

    D[z] = 17.88

Step-by-step explanation:

In general

V([x+y] = V[x] + V[y] +2Cov[xy]

how in this problem Cov[XY] = 0, then

V[x+y] = V[x] + V[y]

Also we must use this properti of the variance  

V[ax+b] = a^{2}V[x]

and remember that

standard desviation = \sqrt{Var[x]}

a) z = 35-10x

   Var[z] = 10^{2} Var[x] = 100*16 = 1600

   D[z] = \sqrt{1600} = 40

b) z = 12x -5

   Var[z] = 12^{2} Var[x] = 144*16 = 2304

   D[z] = \sqrt{2304} = 48

c) z = x + y

   Var[z] =  Var[x+y] = Var[x] + Var[y] = 16 + 64 = 80

   D[z] = \sqrt{80} = 8.94  

d) z = x - y

   Var[z] =  Var[x-y] = Var[x] + Var[y] = 16 + 64 = 80

   D[z] = \sqrt{80} = 8.94

e) z = -2x + 2y

   Var[z] = 4Var[x] + 4Var[y] = 4*16 + 4*64 = 320

  D[z] = \sqrt{320} = 17.88

   

7 0
3 years ago
The perimeter of this figure Is 42.5<br><br>Find the area of the figure.​
Murljashka [212]

Answer:

Formula - 42.5/4^2\\

(perimeter/4^2)

42.5/4 = 10.625

10.625^2 = 112.890625

Answer - 112.890625

5 0
2 years ago
Write as a Improper fraction
Likurg_2 [28]
38/9
46/7
44/5
11/4
19/2
11/6
23/3
41/8
52/7
34/5
8 0
2 years ago
Read 2 more answers
If 28% of students in College are near-sighted, the probability that in a randomly chosen group of 20 College students, exactly
cluponka [151]

Answer: (D) 16%

Step-by-step explanation:

Binomial probability formula :-

P(x)=^nC_xp^x(1-p)^n-x, where n is the sample size , p is population proportion and P(x) is the probability of getting success in x trial.

Given : The proportion of students in College are near-sighted : p= 0.28

Sample size : n= 20

Then, the the probability that in a randomly chosen group of 20 College students, exactly 4 are near-sighted is given by :_

P(x=4)=^{20}C_4(0.28)^4(1-0.28)^{20-4}\\\\=\dfrac{20!}{4!16!}(0.28)^4(0.72)^{16}\\\\=0.155326604912\approx0.16\%

Hence, the probability that in a randomly chosen group of 20 College students, exactly 4 are near-sighted is closest to 16%.

7 0
3 years ago
Solve the system of equations by substitution. <br> x+y=18<br> y=8x
Naddika [18.5K]

Answer:

x=2, y=16

Step-by-step explanation:

x+y=18

y=8x

x+8x=18

x=2

y=16

3 0
3 years ago
Read 2 more answers
Other questions:
  • Alexander is going to an amusement park. The price of admission into the park is $15, and once he is inside the park, he will ha
    14·1 answer
  • If 1 hour is 60 minutes then what will 5 hours equal
    6·2 answers
  • Which of the following expressions are equivalent ?
    13·1 answer
  • Kory bought 2 boxes of candy and 1 soda at the movies and paid $8.00.christopher bought 1 box of candy and 1 soda and paid $5.50
    10·1 answer
  • The table below shows two equations:
    10·2 answers
  • Im giving brainlyest to first CORECT answer and 30points QUICKKKKKK
    8·1 answer
  • Jane has 6/7 of a yard of ribbon. She uses 3/4 of a yard to trim an ornament. how much ribbon is left?
    8·2 answers
  • The bag contains green yellow and orange marbles. The ratio of green marbles to yellow marbles is 2 to 5. The ratio of yellow ma
    6·1 answer
  • Riley rakes 1/6 of a lawn in 2/3 hour. How many lawns can Riley rake per hour?
    9·1 answer
  • 2+2=?<br><br>help I don't know​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!