What is 11.8 as a mixed number?

1 Answer

Answer :

It is 11 and 4/5 as a mixed number in its simplest form

Related questions

Description : Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture? A) 8 : 5 B) 7 : 5 C) 5 : 11 D) 4 : 9 E) 5 : 7

Last Answer : Answer: E Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg Total brass = 30+40 = 70 kg So copper in mixture is (50+70) – 70 = 50 kg So copper to brass = 50 : 70=5:7

Description : What is 34 over 11 as a mixed number?

Last Answer : It is 3 and 1/11 as a mixed number

Description : What is a mixed number of 11 over 4?

Last Answer : It is 2 3/4.

Description : How do you write 11.4 as a mixed number in simplest form?

Last Answer : It is 11 and 2/5 as a mixed number in its simplest form

Description : Tickets numbered 1 to 37 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 10? A) 11/37 B) 37/11 C) 12/37 D) 37/12

Last Answer : Answer: A) Here, S = {1, 2, 3, 4, ...., 36,37}. Let E = event of getting a multiple of 4 or 10= {4,8,12,16,20,24,28,32,36,10, 30}. P(E) = n(E)/n(S) = 11/37

Description : When 22.4 litres of `H_(2)(g)` is mixed with 11.2 litres of `CI_(2)(g)`, eacg at STP, the moles of HCI(g) formed is equal to :-

Last Answer : When 22.4 litres of `H_(2)(g)` is mixed with 11.2 litres of `CI_(2)(g)`, eacg at STP, the moles of HCI(g ... ) C. 1 mol of HCI (g) D. 2 mol of HCI (g)

Description : Saturated steam at 1 atm is discharged from a turbine at 1200 kg/h. Superheated steam at 300 0C  and 1 atm is needed as a feed to a heat exchanger. To produce it, the turbine discharge ... amount of superheated steam at 300 0C produced and the volumetric flow rate of the  400 0C steam.

Last Answer : Solution 1. Mass balance of water 1200 + m1 = m2 ………………………………………… (1) …………………….1 mark 2. Energy balance (1200 kg/h)(2676 kJ/kg) + m1(3278 kJ/kg)

Description : Two vessels contains equal quantity of solution contains milk and water in the ratio of 7:2 and 4:5 respectively. Now the solutions are mixed with each other then find the ratio of milk and water in the final solution? A) 11:7 B) 11:6 C) 11:5 D) 11:9 E) None of these

Last Answer : Answer: A milk = 7/9 and water = 2/9 – in 1st vessel milk = 4/9 and water = 5/9 – in 2nd vessel (7/9 + 4/9)/ (2/9 + 5/9) = 11:7