68 times 157?

Description : Hyperargininemia, a defect in urea synthesis develops due to deficiency of the enzyme: (A) Ornithine transcarbamoylase (B) Argininosuccinase (C) Arginase (D) Argininosuccinate synthetase ENZYMES 157

Last Answer : Answer : B

According to 1998 census the density of population per Sq. Km in FATA is: A. 98 Persons per Sq. Km B. 115 Persons per Sq. Km C. 132 Persons per Sq. Km D. 157 Persons per Sq. Km

Description : According to 1998 census the density of population per Sq. Km in FATA is: A. 98 Persons per Sq. Km B. 115 Persons per Sq. Km C. 132 Persons per Sq. Km D. 157 Persons per Sq. Km

Last Answer : ANSWER: C

The density of population in Pakistan according to 1998 census was: A. 120 Persons per sq. km B. 132 Persons per sq. km C. 157 Persons per sq. km D. 164 Persons per sq. km

Description : The density of population in Pakistan according to 1998 census was: A. 120 Persons per sq. km B. 132 Persons per sq. km C. 157 Persons per sq. km D. 164 Persons per sq. km

Last Answer : ANSWER: D

According to 1998 census the density of population per Sq. Km in FATA is: A. 98 Persons per Sq. Km B. 115 Persons per Sq. Km C. 132 Persons per Sq. Km D. 157 Persons per Sq. Km

Description : According to 1998 census the density of population per Sq. Km in FATA is: A. 98 Persons per Sq. Km B. 115 Persons per Sq. Km C. 132 Persons per Sq. Km D. 157 Persons per Sq. Km

Last Answer : ANSWER: C

The density of population in Pakistan according to 1998 census was: A. 120 Persons per sq. km B. 132 Persons per sq. km C. 157 Persons per sq. km D. 164 Persons per sq. km

Description : The density of population in Pakistan according to 1998 census was: A. 120 Persons per sq. km B. 132 Persons per sq. km C. 157 Persons per sq. km D. 164 Persons per sq. km

Last Answer : ANSWER: D

The value of Chezy’s constant C = 157.6/(1.81 + K/ ) is used in A. Chezy’s formula B. Bazin’s formula C. Kutter’s formula D. Manning’s formula

Description : The value of Chezy’s constant C = 157.6/(1.81 + K/ ) is used in A. Chezy’s formula B. Bazin’s formula C. Kutter’s formula D. Manning’s formula

Last Answer : ANS: B

Calculate the natural frequency of transverse vibrations if the static deflection is 0.01mm. a) 157.6 b) 144.8 c) 173.2 d) 154.1

Description : Calculate the natural frequency of transverse vibrations if the static deflection is 0.01mm. a) 157.6 b) 144.8 c) 173.2 d) 154.1

Last Answer : a) 157.6

A cable of core radius 1.25cm and impregnated paper insulation of thickness 2.13cm and relative permittivity 3.5. Compute the capacitance of the cable/km(in nF) a) 195.7 b) 179.5 c) 157.9 d) 197.5

Description : A cable of core radius 1.25cm and impregnated paper insulation of thickness 2.13cm and relative permittivity 3.5. Compute the capacitance of the cable/km(in nF) a) 195.7 b) 179.5 c) 157.9 d) 197.5

Last Answer : a) 195.7

Locomotive boiler has a) 137 fire tubes and 44 superheated tubes b) 147 fire tubes and 34 superheated tubes c) 157 fire tubes and 24 superheated tubes d) 167 fire tubes and 14 superheated tubes

Description : Locomotive boiler has a) 137 fire tubes and 44 superheated tubes b) 147 fire tubes and 34 superheated tubes c) 157 fire tubes and 24 superheated tubes d) 167 fire tubes and 14 superheated tubes

Last Answer : c) 157 fire tubes and 24 superheated tubes

Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Description : Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Last Answer : 15.7 BTU

4×(√3 + √4)2 + 6(√5 +√6)2 -3(√2 + √3)2 =? a) 167 b) 123 c) 157 d) 153 e) 149

Description : 4×(√3 + √4)2 + 6(√5 +√6)2 -3(√2 + √3)2 =? a) 167 b) 123 c) 157 d) 153 e) 149

Last Answer : ? =4(3+4+2√12) + 6(5+6+2√30)-3(2+3+2√6) =4(7+2√12)+6(11+2√30)-3(5+2√6) =28 + 8√12 + 66 + 12√30 – 15 - 6√6 =(28+66-15) + (8√12+12√30-6√6) =79 + (8√4 × 3 + 12√30 - 6√6) =79+16√3 + 12√30 - 6√6 =79 + 16 ×1.7 + 12 × 5.4 – 6 ×2.4 =79+27.2 + 64.8-14.4 = 156.6 = 157 Answer: c)

The reflex angle between the hands of a clock at 10.25 is a) 197.5 b) 167.5 c) 157.5 d) 187.5

Description : The reflex angle between the hands of a clock at 10.25 is a) 197.5 b) 167.5 c) 157.5 d) 187.5

Last Answer : Answer : b

5, 14 , 41 , 86 , ? (a) 149 (b) 123 (c) 157 (d) 131

Description : 5, 14 , 41 , 86 , ? (a) 149 (b) 123 (c) 157 (d) 131

Last Answer : The difference between the numbers is the multiple of 9 with odd numbers .

What is the ratio between money spent by company Q on tax and money spent by P on salary? a) 23:121 b) 33:144 c) 47:144 d) 144:157 e) None of these

Description : What is the ratio between money spent by company Q on tax and money spent by P on salary? a) 23:121 b) 33:144 c) 47:144 d) 144:157 e) None of these

Last Answer : e) None of these

Total literate population of City F is approximately what percentage of total illiterate population of City A ? 1) 137.5% 2) 142.5% 3) 147.5% 4) 152.5% 5) 157.5%

Description : Total literate population of City F is approximately what percentage of total illiterate population of City A ? 1) 137.5% 2) 142.5% 3) 147.5% 4) 152.5% 5) 157.5%

Last Answer : 4) 152.5%

Convert – (255)10 = (?)16 = (?)8 (157)10 = (?)BCD = (?) Excess3

Description : Convert – (255)10 = (?)16 = (?)8 (157)10 = (?)BCD = (?) Excess3

Last Answer : Solution:

How many times does 9 go into 68?

Description : How many times does 9 go into 68?

Last Answer : 7 times with a remainder of 5

How many times can for go into 68?

Description : How many times can for go into 68?

Last Answer : Four goes into 68 17 times.

How many times does 68 go into 612?

Description : How many times does 68 go into 612?

Last Answer : Exactly 9 times

How many times does 18 into 68?

Description : How many times does 18 into 68?

Last Answer : 18 + 18 + 18 = 36 + 18 = 5468 - 54 = 143 times with 14 remaining

How many times dose 8 go into 68?

Description : How many times dose 8 go into 68?

Last Answer : Feel Free to Answer

What would be a more reasonable estimate for 13.68 and times?

Description : What would be a more reasonable estimate for 13.68 and times?

Last Answer : What is the answer ?

How many times can 17 go into 68 without going over?

Description : How many times can 17 go into 68 without going over?

Last Answer : four17 + 17 = 3434 + 34 = 68

How many times can 17 go into 68 without going over?

Description : How many times can 17 go into 68 without going over?

Last Answer : four17 + 17 = 3434 + 34 = 68

How many times does 14 go into 68?

Description : How many times does 14 go into 68?

Last Answer : Four times... with 12 remaining.

How many times does 2 go in 68?

Description : How many times does 2 go in 68?

Last Answer : Exactly 34 times

How many times does 4 go into 68?

Description : How many times does 4 go into 68?

Last Answer : 14

Madhan was 3 times as old as his daughter 10 years ago. After 10 years, Madhan will be twice as old as his daughter. Find out the present age of Madhan. a) 68 years b) 70 years c) 40 years d) 55 years e) 80 years

Description : Madhan was 3 times as old as his daughter 10 years ago. After 10 years, Madhan will be twice as old as his daughter. Find out the present age of Madhan. a) 68 years b) 70 years c) 40 years d) 55 years e) 80 years

Last Answer : b Let age of the daughter before 10 years = x Then, age of Madhan before 10 years = 3x After 10 years, Madhan will be twice as old as his daughter. ⇒ 3x + 20 = 2( x+20) ⇒ x = 20 ⇒ Present age of Madhan = 3x + 10 = 3× 20+ 10 = 70 years.

A computer program selects an integer in the set {k : 1 ≤ k ≤ 10,00,000} at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k = 1 appears in the printout atleast once ? (A) 0.5 (B) 0.704 (C) 0.632121 (D) 0.68

Description : A computer program selects an integer in the set {k : 1 ≤ k ≤ 10,00,000} at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k = 1 appears in the printout atleast once ? (A) 0.5 (B) 0.704 (C) 0.632121 (D) 0.68

Last Answer : (C) 0.632121

Do you think a 2004 Ex Honda Element with 68,500miles priced at $8,000 is a good buy?

Description : Do you think a 2004 Ex Honda Element with 68,500miles priced at $8,000 is a good buy?

Last Answer : That element machine holds together really well. Those things will rack up some mileage.

Thoughts on Hillary Clinton's comment about Bobby Kennedy assasination during the primary in '68?

Description : Thoughts on Hillary Clinton's comment about Bobby Kennedy assasination during the primary in '68?

Last Answer : Either she is insinuating that Obama might be assassinated so she should stay in the race (which sounds like a threat to me!) or she was just trying to reference a point in history to show that it is too ... like this, so I think it is probably the latter option. But I really don't know. Go Obama!

What is the total acceleration of a point on the rim of a flywheel 4 mts in diameter 0.6 see after it has started from rest if it accelarates uniformly from rest to 2000 rpm in 20 secs. a.68.9 mts/sec2 b.102.6 mts/sec2 c.78.94 mts/sec2 d.107 dynes e.33.2 mts/sec2

Description : What is the total acceleration of a point on the rim of a flywheel 4 mts in diameter 0.6 see after it has started from rest if it accelarates uniformly from rest to 2000 rpm in 20 secs. a.68.9 mts/sec2 b.102.6 mts/sec2 c.78.94 mts/sec2 d.107 dynes e.33.2 mts/sec2

Last Answer : c. 78.94 mts/sec2

Work done by a body of mass 100 kg moving along a circular path of radius 2 m with a uniform velocity of 10 m/sec is a.5000 J b.2500 J c.509.68 J d.None of the above e.Tapered bearing

Description : Work done by a body of mass 100 kg moving along a circular path of radius 2 m with a uniform velocity of 10 m/sec is a.5000 J b.2500 J c.509.68 J d.None of the above e.Tapered bearing

Last Answer : d. None of the above

Look at this sequence; I6, 06, 68, 88, X, 98. What does the X represent? -Riddles

Description : Look at this sequence; I6, 06, 68, 88, X, 98. What does the X represent? -Riddles

Last Answer : The answer is 87, in order to solve this you need to look at the numbers upside down. The sequence comes out to be 86, ?, 88, 89, 90, 91; therefore the ? is 87.

16 06 68 88 ? 98. Fill in the ? -Riddles

Description : 16 06 68 88 ? 98. Fill in the ? -Riddles

Last Answer : Look at the number, upside down. The order is 86, 87, 88, 89, 90, 91… The answer is 87.

The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Last Answer : NEED ANSWER

The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Last Answer : According to question find the median of the data

What is the value of 68?

Description : What is the value of 68?

Last Answer : me

68+ what equals 144?

Description : 68+ what equals 144?

Last Answer : 144-68= 76. so your answer is 68+76=144

Is 17/20 0.68 or 0.85?

Description : Is 17/20 0.68 or 0.85?

Last Answer : Nevermind I figured out it is 0.85

Using the following thermochemical data. `C(S)+O_(2)(g)rarr CO_(2)(g), Delta H=94.0` Kcal `H_(2)(g)+1//2O_(2)(g)rarr H_(2)O(l), Delta H=-68.0` Kcal `C

Description : Using the following thermochemical data. `C(S)+O_(2)(g)rarr CO_(2)(g), Delta H=94.0` Kcal `H_(2)(g)+1//2O_(2)(g)rarr H_(2)O(l), Delta H=-68.0` Kcal `C

Last Answer : Using the following thermochemical data. `C(S)+O_(2)(g)rarr CO_(2)(g), Delta H=94.0` Kcal `H_(2)(g)+1// ... .0` Kcal C. `-114.0` Kcal D. `+114.0` Kcal

The heat of combustion of `CH_(4(g)), C_((g))` and `H_(2(g))` at `25^(@)C` are `-212.4` K cal, -94.0 K cal and -68.4 K cal respectively, the heat of f

Description : The heat of combustion of `CH_(4(g)), C_((g))` and `H_(2(g))` at `25^(@)C` are `-212.4` K cal, -94.0 K cal and -68.4 K cal respectively, the heat of f

Last Answer : The heat of combustion of `CH_(4(g)), C_((g))` and `H_(2(g))` at `25^(@)C` are `-212.4` K cal, -94.0 ... .4` K cal C. `-375.2` K cal D. `+212.8` K cal

68 times 157?

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