A concrete plant uses the following proportions per m3 of concrete: cement 310 kg; water 180 kg; fine aggregate 750 kg and coarse aggregate 1080 kg both in SSD conditions. If the fine aggregate in the field has 4% excess water find the revised quantity of water. cehgg

150 kg

Step-by-step explanation:

Given that

cement = 310 kg  

Fine aggregate = 750 kg

coarse aggregate = 1080 kg

mass of water = 180 kg

given that aggregate in field is 4% excess water

Determine the revised quantity of water

mass of concrete per m^3 = mass of cement + mass of water + mass of fine aggregate + mass of coarse aggregate

therefore mass of concrete = 310 + 180  + 750 + 1080 =  2320 kg

given that fine aggregate in field has 4% excess water this simply means it was void of air in the concrete the new weight = (1.04 * 750) = 780

Hence the revised quantity of water ( x )

= 310 + x + 780 + 1080 = 2320

therefore x = 150 kg

oof cheaters cheat because they’re lazy that’s why i cheat xd

: )

step-by-step explanation:

first, we define the variables:

x: number of years after 1950

f (x): amount of vinyl sold.

then, with the variables defined, we have:

68594 vinyl records were sold in 1958 > f (8) = 68594

91299 vinyl records were sold in 1961 > f (11) = 91299

38720 vinyl records were sold in 1952 > f (2) = 38720

161743 vinyl records were sold in 1967 > f (17) = 161743

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step-by-step explanation: