Answer:
Her usual rate is 0.8333 miles per hour.
Step-by-step explanation:
The velocity formula is given by:
![v = \frac{d}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D)
In which d is the distance and t is the time.
She completed the first half of the walk 1 mi/h faster than usual
Her usual rate is v. 1mph faster is v + 1.
The second half of the walk 2 mi/h slower than the first half.
The first half is v + 1.
2mph slower is v + 1 - 2 = v - 1. Then
Total rate:
7.2 hours, and 6 miles. So
One half is v+1 and the other is v - 1. This is why each is multiplied by 0.5.
![0.5(v + 1) + 0.5(v - 1) = \frac{6}{7.2}](https://tex.z-dn.net/?f=0.5%28v%20%2B%201%29%20%2B%200.5%28v%20-%201%29%20%3D%20%5Cfrac%7B6%7D%7B7.2%7D)
![0.5v + 0.5 + 0.5v - 0.5 = 0.8333](https://tex.z-dn.net/?f=0.5v%20%2B%200.5%20%2B%200.5v%20-%200.5%20%3D%200.8333)
![v = 0.8333](https://tex.z-dn.net/?f=v%20%3D%200.8333)
Her usual rate is 0.8333 miles per hour.
Answer:104ft^3
Step-by-step explanation:
volume=length x width x height
Volume=13 x 4 x 2
Volume=104ft^3
Answer:
It's the second one. Press 'Ask For Help' if you need more help on problems
Answer:
![c=0.4\ J/g^{\circ} C](https://tex.z-dn.net/?f=c%3D0.4%5C%20J%2Fg%5E%7B%5Ccirc%7D%20C)
Step-by-step explanation:
Given that,
Mass of a sample, m = 12 g
Heat absobed by the sample, Q = 96 J
It was heated from 20°C to 40°C.
We need to find the specific heat of a material. The heat absorbed by a sample is given by :
![Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{96}{12\times (40-20)}\\\\=0.4\ J/g^{\circ} C](https://tex.z-dn.net/?f=Q%3Dmc%5CDelta%20T%5C%5C%5C%5Cc%3D%5Cdfrac%7BQ%7D%7Bm%5CDelta%20T%7D%5C%5C%5C%5Cc%3D%5Cdfrac%7B96%7D%7B12%5Ctimes%20%2840-20%29%7D%5C%5C%5C%5C%3D0.4%5C%20J%2Fg%5E%7B%5Ccirc%7D%20C)
So, the specific heat of the material is
.
Answer:
Step-by-step explanation:
Given that events A1, A2 and A3 form a partiton of the sample space S with probabilities P(A1) = 0.3, P(A2) = 0.5, P(A3) = 0.2.
i.e. A1, A2, and A3 are mutually exclusive and exhaustive
E is an event such that
P(E|A1) = 0.1, P(E|A2) = 0.6, P(E|A3) = 0.8,
![P(E) = P(A_1E)+P(A_2E)+P(A_3E)\\= \Sigma P(E/A_1) P(A_1) \\= 0.1(0.3)+0.5(0.6)+0.8(0.2)\\= 0.03+0.3+0.16\\= 0.49](https://tex.z-dn.net/?f=P%28E%29%20%3D%20P%28A_1E%29%2BP%28A_2E%29%2BP%28A_3E%29%5C%5C%3D%20%5CSigma%20P%28E%2FA_1%29%20P%28A_1%29%20%5C%5C%3D%200.1%280.3%29%2B0.5%280.6%29%2B0.8%280.2%29%5C%5C%3D%200.03%2B0.3%2B0.16%5C%5C%3D%200.49)
![P(A_1/E) = P(A_1E)/P(E) = \frac{0.3(0.1)}{0.49} \\=0.061224](https://tex.z-dn.net/?f=P%28A_1%2FE%29%20%3D%20P%28A_1E%29%2FP%28E%29%20%3D%20%5Cfrac%7B0.3%280.1%29%7D%7B0.49%7D%20%5C%5C%3D0.061224)
![P(A_2/E) = P(A_2E)/P(E) = \frac{0.5)(0.6)}{0.49} \\=0.61224](https://tex.z-dn.net/?f=P%28A_2%2FE%29%20%3D%20P%28A_2E%29%2FP%28E%29%20%3D%20%5Cfrac%7B0.5%29%280.6%29%7D%7B0.49%7D%20%5C%5C%3D0.61224)
![P(A_3/E) = P(A_3E)/P(E) = \frac{0.2)(0.8)}{0.49} \\=0.3265](https://tex.z-dn.net/?f=P%28A_3%2FE%29%20%3D%20P%28A_3E%29%2FP%28E%29%20%3D%20%5Cfrac%7B0.2%29%280.8%29%7D%7B0.49%7D%20%5C%5C%3D0.3265)