Vraag 33

Voorbeelden van een juist antwoord zijn:

•Vooris100:15=6{,}7~\text{L}100:15=6{,}7~\text{ L}100:15=6{,}7~100:15=6{,}7~100:15=6{,}7~100:15=6{,}7~100:15=6{,}7~100:15=6{,}7~100:15=6{,}7~100:15=6{,}7~100:15=6{,}7~100:15=6{,}7~$100: 15=6{,}7 \mathrm{~L}benzine nodig. De dichtheid van benzine is (volgens Binas-tabel 16)0{,}72~\text{g/cm}^3=0{,}72~\text{kg/L}0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L$0{,}72 \mathrm{~g} / \mathrm{cm}^{3}=0{,}72 \mathrm{~kg} / \mathrm{L}. Dus vooris6{,}7\times0{,}72=4{,}8~\text{kg}6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~6{,}7\times0{,}72=4{,}8~k$6{,}7 \times 0{,}72=4{,}8 \mathrm{~kg}benzine nodig (dit is minder danijzerpoeder).

•De dichtheid van benzine is (volgens Binas-tabel 16 )0{,}72~\text{g/cm}^3=0{,}72~\text{kg/L}0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~0{,}72~\text{g/cm}^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L0{,}72~^3=0{,}72~kg/L$0{,}72 \mathrm{~g} / \mathrm{cm}^{3}= 0{,}72 \mathrm{~kg} / \mathrm{L}. Dus per km is dat0{,}72:15=0{,}048~\text{kg}0{,}72:15=0{,}048~\text{lkg}0{,}72:15=0{,}048~\text{lg}0{,}72:15=0{,}048~0{,}72:15=0{,}048~0{,}72:15=0{,}048~0{,}72:15=0{,}048~0{,}72:15=0{,}048~0{,}72:15=0{,}048~0{,}72:15=0{,}048~0{,}72:15=0{,}048~0{,}72:15=0{,}048~0{,}72:15=0{,}048~0{,}72:15=0{,}048~k$0{,}72: 15=0{,}048 \mathrm{~kg}benzine. Per km is40\,(\text{kg}):100\,(\text{km})=0{,}4~\text{kg}40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~40\,(\text{kg}):100\,(\text{km})=0{,}4~k40\,(\text{kg}):100\,(\text{km})=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,()=0{,}4~kg40\,(\text{kg}):100\,(km)=0{,}4~kg40\,(\text{kg}):100(km)=0{,}4~kg40\,(\text{kg}):100(km)=0{,}4~kg40\,(\text{kg}):100(km)=0{,}4~kg40\,(\text{kg}):100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,():100(~km)=0{,}4~kg40\,(k):100(~km)=0{,}4~kg40\,(kg):100(~km)=0{,}4~kg40\,(~kg):100(~km)=0{,}4~kg40(~kg):100(~km)=0{,}4~kg40(~kg):100(~km)=0{,}4~kg$40(\mathrm{~kg}): 100(\mathrm{~km})=0{,}4 \mathrm{~kg}ijzerpoeder nodig (en dit is meer dan0{,}048~\text{kg}0{,}048~0{,}048~0{,}048~0{,}048~0{,}048~0{,}048~0{,}048~0{,}048~0{,}048~0{,}048~0{,}048~k$0{,}048 \mathrm{~kg}benzine).

of

Opmerking

De significantie bij deze berekening niet beoordelen.