What is equivalent to 154?

What is equivalent to 154?
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154 is equivalent to 154.0 or 154/1
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What is equivalent to 154?

Description : What is equivalent to 154?

Last Answer : 154 is equivalent to 154.0 or 154/1

The octal equivalent of 1100101.001010 is ______ a) 624.12 b) 145.12 c) 154.12 d) 145.21

Description : The octal equivalent of 1100101.001010 is ______ a) 624.12 b) 145.12 c) 154.12 d) 145.21

Last Answer : Answer: b Explanation: The octal equivalent is obtained by grouping the numbers into three, from right to left before decimal and from right to left after the decimal place. Here, i.e. 145.12 is the octal equivalent of the number

4° F will be equivalent to (a) 77° X (b) 113.3° X (c) 128.6° X (d) 154.4° X

Description : 4° F will be equivalent to (a) 77° X (b) 113.3° X (c) 128.6° X (d) 154.4° X

Last Answer : Ans:(c)

Why would someone pay $3,154 on interest on a new car loan when they could buy a decent used one for under that amount without a loan (or interest)?

Description : Why would someone pay $3,154 on interest on a new car loan when they could buy a decent used one for under that amount without a loan (or interest)?

Last Answer : I buy new cars about every eight or ten years. I had my share of used vehicles when I was younger and spent a lot more than 3,000 on upkeep, repairs, replacing parts, etc. I have no idea how to work on ... tires, and brakes, then I buy a new one. I have never spent over 18,000 on a new car though.

Given that the area of a figure varies as square of mean ordinate. The area of the figure is 154 sq. cm when the mean ordinate is 7 cm. The area of the figure when mean ordinate is 10.5 cm will be a.364.9 cm b.364.5 cm c.360.6 cm d.346.5 cm e.356.3 cm

Description : Given that the area of a figure varies as square of mean ordinate. The area of the figure is 154 sq. cm when the mean ordinate is 7 cm. The area of the figure when mean ordinate is 10.5 cm will be a.364.9 cm b.364.5 cm c.360.6 cm d.346.5 cm e.356.3 cm

Last Answer : d. 346.5 cm

the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Description : the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Last Answer : T.S.A = 3*154 = 462 cm² C.S.A = 154 cm² C.S.A = 2πrh T.S.A = 2πr(r+h) Now, In T.S.A = 2πrr + 2πrh 462 = 2πrr + 2πrh 462 = 2*22/7*r*r + 154 462 - 154 = 2*22/7*r*r 308*7/2*22 = r*r 49 = r*r R = 7 cm ... 7*h 154/44 = h 7/2 =h H = 3.5 cm or 7/2 cm Now volume = πrrh = 22/7 * 7* 7 *7/2 = 11*49 = 539 cm³

A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.

The curved surface area of a cylinder is 154 cm2. -Maths 9th

Description : The curved surface area of a cylinder is 154 cm2. -Maths 9th

Last Answer : Since curved surface area of a cyclinder = 154 cm2 [given] Total surface area of cyclinder = 3 curved surface area 2πrh + 2πr2 = 3 154 154 + 2πr2 = 462 2πr2 = 462 - 154 = 308 r2 = 308 7 / 2 22 = 49 ... 154 / 44 = 3.5 cm ∴ Volume of cyclinder = πr2h = 22 / 7 7 7 3.5 = 539 cm3

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th

Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.

The curved surface area of a cylinder is 154 cm2. -Maths 9th

Description : The curved surface area of a cylinder is 154 cm2. -Maths 9th

Last Answer : Since curved surface area of a cyclinder = 154 cm2 [given] Total surface area of cyclinder = 3 curved surface area 2πrh + 2πr2 = 3 154 154 + 2πr2 = 462 2πr2 = 462 - 154 = 308 r2 = 308 7 / 2 22 = 49 ... 154 / 44 = 3.5 cm ∴ Volume of cyclinder = πr2h = 22 / 7 7 7 3.5 = 539 cm3

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

Description : Find the radius of a sphere whose surface area is 154 cm square. -Maths 9th

Last Answer : Let 'r' be the radius of sphere Surface area of sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 154 = 4 x 22/7 x r2 ⇒ r 2 = 154 x 7/4 x 22 = 49/4 ⇒ r = 7/2 cm = 3.5 cm

Find the volume of a sphere whose surface area is 154 cm sq. -Maths 9th

Description : Find the volume of a sphere whose surface area is 154 cm sq. -Maths 9th

Last Answer : Let r cm be the radius of sphere. Surface area of the sphere = 4 πr2 ⇒ 154 = 4 πr2 ⇒ 4 x 22/7 x r2 = 154 r 2 = 154 x 7/4 x 22 = 72/22 ⇒ r = 7/2 Volume of sphere = 4/3 πr3 = 4/3 x 22/7 x 7/2 x 7/2 x 7/2 cm3 = 539/3 cm3 = 179.2/3 cm3

Reread pp 152-154 through the quote shown below. List everything you learn about Mr. Chris from this part of the text as a person-his character in other words. One way an author helps you learn about a character is through what others say about him?

Description : Reread pp 152-154 through the quote shown below. List everything you learn about Mr. Chris from this part of the text as a person-his character in other words. One way an author helps you learn about a character is through what others say about him?

Last Answer : This is about Miracles on Maple Hill. The quote: " This man looks like a tree" and "He seemed to stand with his legs planted in the ground.

The area of a circle (A) varies directly as the square of its radius. The area of the circle of radius 7 cm is 154 `cm^(2)`. What is the area of the c

Description : The area of a circle (A) varies directly as the square of its radius. The area of the circle of radius 7 cm is 154 `cm^(2)`. What is the area of the c

Last Answer : The area of a circle (A) varies directly as the square of its radius. The area of the circle of ... What is the area of the circle of radius 35 cm?

How do you write 154 in word form?

Description : How do you write 154 in word form?

Last Answer : one hundred and fifty-four

How many times can two go into 154?

Description : How many times can two go into 154?

Last Answer : Exactly 77 times

How many times does 10 go into 154?

Description : How many times does 10 go into 154?

Last Answer : 15 R 4

What is thr hcf of 84 154 and 182?

Description : What is thr hcf of 84 154 and 182?

Last Answer : It is: 14

What is the name for a 154 angle?

Description : What is the name for a 154 angle?

Last Answer : abctue

What is the name for a 154 angle?

Description : What is the name for a 154 angle?

Last Answer : abctue

232+154?

Description : 232+154?

Last Answer : 386

1 molecule of palmitic acid on total oxidation to CO2 will yield molecules of ATP (as high energy bonds): (A) 129 (B) 154 (C) 83 (D) 25

Description : 1 molecule of palmitic acid on total oxidation to CO2 will yield molecules of ATP (as high energy bonds): (A) 129 (B) 154 (C) 83 (D) 25

Last Answer : Answer : A

Total Population of Livestock in Pakistan is: A. 127.5 Million B. 130 Million C. 154.7 Million D. 159.3 Million

Description : Total Population of Livestock in Pakistan is: A. 127.5 Million B. 130 Million C. 154.7 Million D. 159.3 Million

Last Answer : ANSWER: C

Total Population of Livestock in Pakistan is: A. 127.5 Million B. 130 Million C. 154.7 Million D. 159.3 Million

Description : Total Population of Livestock in Pakistan is: A. 127.5 Million B. 130 Million C. 154.7 Million D. 159.3 Million

Last Answer : ANSWER: C

The appointment of a Governor in a state is made as per the provision in the Constitution under article. (1) 153 (2) 154 (3) 155 (4) 156

Description : The appointment of a Governor in a state is made as per the provision in the Constitution under article. (1) 153 (2) 154 (3) 155 (4) 156

Last Answer : (3) 155 Explanation: Appointment of the Governor of Indian state is described under Article 155 of the Indian constitution.

Which article of Indian Constitution declares Devnagri Hindi as an official language of India? (1) Article 343 (2) Article 348 (3) Article 154 (4) Article 156

Description : Which article of Indian Constitution declares Devnagri Hindi as an official language of India? (1) Article 343 (2) Article 348 (3) Article 154 (4) Article 156

Last Answer : (1) Article 343 Explanation: Article 343 declares Devnagri Hindi as an official language of India. Article 348 is about language used in the Supreme Court and High Courts. Article 154 deals with executive power of state. Article 156 declares term of office of Governor.

When a voltage of 110 VDC is applied to the illustrated circuit with a resistance of 237 ohms the current will be________. EL-0018 A. .464 amps B. 1.464 amps C. 2.154 amps D. 3.47 amps

Description : When a voltage of 110 VDC is applied to the illustrated circuit with a resistance of 237 ohms the current will be________. EL-0018 A. .464 amps B. 1.464 amps C. 2.154 amps D. 3.47 amps

Last Answer : Answer: A

When a voltage of 115 VDC is applied to the illustrated circuit with a resistance of 17.8 ohms the current will be _______. EL-0018 A. 0.154 amps B. 2.755 amps C. 6.46 amps D. 0.1328 amps

Description : When a voltage of 115 VDC is applied to the illustrated circuit with a resistance of 17.8 ohms the current will be _______. EL-0018 A. 0.154 amps B. 2.755 amps C. 6.46 amps D. 0.1328 amps

Last Answer : Answer: C

200 mesh sieve size corresponds to __________ microns. (A) 24 (B) 74 (C) 154 (D) 200

Description : 200 mesh sieve size corresponds to __________ microns. (A) 24 (B) 74 (C) 154 (D) 200

Last Answer : (B) 74

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

In a circle of radius 14 cm, an arc subtends an angle of 45 O at the centre, then the area of the sector is (a) 71 cm 2(b) 76 cm 2 (c) 77 cm 2 (d) 154 cm 2

Description : In a circle of radius 14 cm, an arc subtends an angle of 45 O at the centre, then the area of the sector is (a) 71 cm 2(b) 76 cm 2 (c) 77 cm 2 (d) 154 cm 2

Last Answer : (c) 77 cm 2

1m3 of an ideal gas at 500 K and 1000 kPa expands reversibly to 5 times its initial volume in an insulated container. If the specific heat capacity (at constant pressure) of the gas is 21 J/mole . K, the final temperature will be (A) 35 K (B) 174 K (C) 274 K (D) 154 K

Description : 1m3 of an ideal gas at 500 K and 1000 kPa expands reversibly to 5 times its initial volume in an insulated container. If the specific heat capacity (at constant pressure) of the gas is 21 J/mole . K, the final temperature will be (A) 35 K (B) 174 K (C) 274 K (D) 154 K

Last Answer : (C) 274 K

187.5, 182, 171, ?, 132.5, 105, 72 a) 145.5 b) 155.5 c) 154.5 d) 144.5 e) None of these

Description : 187.5, 182, 171, ?, 132.5, 105, 72 a) 145.5 b) 155.5 c) 154.5 d) 144.5 e) None of these

Last Answer : The pattern is -5.5, -11, -16.5, -22 and so on. Answer: c)

210, 154, 182, 168, 175,171.5, ? a) 172.75 b) 173.25 c) 172.25 d) 173.75 e) 174.75

Description : 210, 154, 182, 168, 175,171.5, ? a) 172.75 b) 173.25 c) 172.25 d) 173.75 e) 174.75

Last Answer : Starting from the third term, every term is the average of the previous two terms. Hence the next term is (175 + 171.5)/ 2 = 173.25 Answer: b)

The area of a circle is 154 sq cm. Its circumference is a) 44cm b) 88cm c) 154cm d) 176cm e) 164 cm

Description : The area of a circle is 154 sq cm. Its circumference is a) 44cm b) 88cm c) 154cm d) 176cm e) 164 cm

Last Answer : a) 44cm

A horse is tied to a peg hammered at one of the corner of a rectangular grass field of 40 m by 24 m by a rope 14 m long. Over how much area of the field can the horse graze? (1) 154 m2 (2) 308 m2 (3) 240 m2 (4) 480 m2

Description : A horse is tied to a peg hammered at one of the corner of a rectangular grass field of 40 m by 24 m by a rope 14 m long. Over how much area of the field can the horse graze? (1) 154 m2 (2) 308 m2 (3) 240 m2 (4) 480 m2

Last Answer : (1) 154 m2

The number of standard rails (13 m) for one km. BG section single line section is (a) 120 (b) 154* (c) 168 (d) 196

Description : The number of standard rails (13 m) for one km. BG section single line section is (a) 120 (b) 154* (c) 168 (d) 196

Last Answer : (b) 154*

Dr. Linus Carl Pauling is the only person to have won two Nobel Prizes individually for: A. Chemistry in 1954, peace prize in 1962 B. Peace prize in 1954, Chemistry in 1962 C. Physics in 154, Medicine in 1962 D. Medicine in 1954, Physics in 1962

Description : Dr. Linus Carl Pauling is the only person to have won two Nobel Prizes individually for: A. Chemistry in 1954, peace prize in 1962 B. Peace prize in 1954, Chemistry in 1962 C. Physics in 154, Medicine in 1962 D. Medicine in 1954, Physics in 1962

Last Answer : A. Chemistry in 1954, peace prize in 1962

The ratio of spacing (d) for for simple cubic lattice, d(100) :d(110) :d(111) a. 1 : 0.707: 0.669 b. 1: 1.414: 0.577 c. 1: 0.707 : 0.577 d. 1: 0.707 : 1.154

Description : The ratio of spacing (d) for for simple cubic lattice, d(100) :d(110) :d(111) a. 1 : 0.707: 0.669 b. 1: 1.414: 0.577 c. 1: 0.707 : 0.577 d. 1: 0.707 : 1.154

Last Answer : c. 1: 0.707 : 0.577

A bus starts at bus stop and reaches a destination in 4 hours. If it travels first and second half at the speed of 40 km/hr and 50 km/hr respectively then the distance between bus stop and destination is a) 177.7 km b) 154.2 km c) 163.5 km d) 147.3 km

Description : A bus starts at bus stop and reaches a destination in 4 hours. If it travels first and second half at the speed of 40 km/hr and 50 km/hr respectively then the distance between bus stop and destination is a) 177.7 km b) 154.2 km c) 163.5 km d) 147.3 km

Last Answer : A Let the distance between bus stop and destination be X. The total time taken by the bus to cover X = 4 hours Since X/2 by 40km/hr and remaining X/2 by 50km/hr Then by Time = distance/speed, we have ... = 1600 9X = 1600 X = 1600/9 = 177.7 Hence 177.7 km is the required answer.

Equal quantities of 3 mixtures of milk and water are mixed in the ratio 1:3, 2:3 and 3:4.The ratio of water and milk in the new mixture is A) 45:76 B) 151:269 C) 123:154 D) 145:245 E) None of these

Description : Equal quantities of 3 mixtures of milk and water are mixed in the ratio 1:3, 2:3 and 3:4.The ratio of water and milk in the new mixture is A) 45:76 B) 151:269 C) 123:154 D) 145:245 E) None of these

Last Answer : Answer: B Milk = 1/4 : 2/5 :3/7 = 35/140 :56/140 : 60/140 Quantity of milk in new mix = 35+56+60 = 151 Quantity of water in new mix = 140*3 = 420-151 = 269 M:W = 151:269

The average age of a morning class is 108, if the average age of 72 ladies in the class is as same as the total average and the number of gents in the class is 2 more than the ladies in the class, what is the average age of gents in the class? A) 152 B) 108 C) 156 D) 154

Description : The average age of a morning class is 108, if the average age of 72 ladies in the class is as same as the total average and the number of gents in the class is 2 more than the ladies in the class, what is the average age of gents in the class? A) 152 B) 108 C) 156 D) 154 

Last Answer : B) According to question, (72+74)X=72×108+74X 146x-74x=7776 72x=7776 X=7776/72= 108years