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\frac{143}{32}=4.46875
Tirohia ngā hipanga rongoā
Ngā Upane Otinga
4 \frac{ 15 }{ 32 }
Whakareatia te 4 ki te 32, ka 128.
\frac{128+15}{32}
Tāpirihia te 128 ki te 15, ka 143.
\frac{143}{32}
Tauwehe
\frac{11 \cdot 13}{2 ^ {5}} = 4\frac{15}{32} = 4.46875
Pātaitai
Arithmetic
5 raruraru e ōrite ana ki:
4 \frac{ 15 }{ 32 }
Ngā Raru Ōrite mai i te Rapu Tukutuku
15/300
https://www.tiger-algebra.com/drill/15/300/
15/300 Final result : 1 —— = 0.05000 20 Step by step solution : Step 1 : 1 Simplify —— 20 Final result : 1 —— = 0.05000 20 Processing ends successfully
Describe 15/(4,3,2,1) as a vector
https://math.stackexchange.com/questions/1902548/describe-15-4-3-2-1-as-a-vector
It looks as if this non-standard notation is asking to find the vector x of the smallest Euclidean norm so that x\cdot (4,3,2,1) = 15. This is the same as making an orthogonal projection (do you ...
What are the mean and standard deviation of a binomial probability distribution with \displaystyle{n}={4} and \displaystyle{p}=\frac{{15}}{{32}} ?
https://socratic.org/questions/what-are-the-mean-and-standard-deviation-of-a-binomial-probability-distribution--62
Mean \displaystyle={1}\frac{{7}}{{8}} SD \displaystyle={0.998} Explanation: Given- \displaystyle{n}={4} \displaystyle{p}=\frac{{15}}{{32}} \displaystyle{q}={1}-{p}={1}-\frac{{15}}{{32}}=\frac{{17}}{{32}} ...
105/30
https://www.tiger-algebra.com/drill/105/30/
105/30 Final result : 7 — = 3.50000 2 Step by step solution : Step 1 : 7 Simplify — 2 Final result : 7 — = 3.50000 2 Processing ends successfully
115/35
http://www.tiger-algebra.com/drill/115/35/
115/35 Final result : 23 —— = 3.28571 7 Step by step solution : Step 1 : 23 Simplify —— 7 Final result : 23 —— = 3.28571 7 Processing ends successfully
If acceleration is increasing at the constant rate of \displaystyle{2}\frac{{m}}{{s}^{{2}}} . Find the distance travelled in \displaystyle{5}{\sec} ? If initial velocity and acceleration both ...
https://socratic.org/questions/if-acceleration-is-increasing-at-the-constant-rate-of-2-m-s-2-find-the-distance-
After 5 seconds the object will have reached a speed of: \displaystyle{v}={5}{s}\times{2}{m}{/}{s}^{{2}}={10}{m}{/}{s} Explanation: Since the speed goes up from 0 to 10 m/s at a constant rate, ...
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Tohaina
Tārua
Kua tāruatia ki te papatopenga
\frac{128+15}{32}
Whakareatia te 4 ki te 32, ka 128.
\frac{143}{32}
Tāpirihia te 128 ki te 15, ka 143.
Ngā Raru Ōrite
3 \frac{ 3 }{ 7 }
4 \frac{ 15 }{ 32 }
1 \frac{ 1 }{ 2 } +3 \frac{ 4 }{ 5 }
1 \frac{ 1 }{ 2 } -3 \frac{ 4 }{ 5 }
1 \frac{ 1 }{ 2 } \times 3 \frac{ 4 }{ 5 }
1 \frac{ 1 }{ 2 } \div 3 \frac{ 4 }{ 5 }
Hoki ki runga
