Answer:
isnt it just 2.50$ ?
Step-by-step explanation:
1$ + 1.5$ = 2.5$ ?
meat + everything else = total
Answer:
1. a + b + c = 180
2. b + d = 180
3. d = a + c
Step-by-step explanation:
1. a + b + c = 180:
Angles in a triangle add up to 180
2. b + d = 180:
Angles in a straight line equal 180 because they are supplementary angles
3. d = a + c:
ΔACD is an exterior angle, and ΔACB is interior adjacent, hence ∠a and ∠c add up to ∠d
Answer:
10.89
Step-by-step explanation:
Circle:
C = TTD
C = TT(6.6)
C = 20.73451151
C = 20.73
Triangle:
A = bh/2
A = 6.6(3.3)/2 6.6/2=3.3
A = 21.78/2
A = 10.89
Shaded Region = 10.89
Answer:
No, the on-time rate of 74% is not correct.
Solution:
As per the question:
Sample size, n = 60
The proportion of the population, P' = 74% = 0.74
q' = 1 - 0.74 = 0.26
We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.
Now,
The proportion of the given sample, p = 
Therefore, the probability is given by:
![P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]](https://tex.z-dn.net/?f=P%28p%5Cleq%200.634%29%20%3D%20%5B%5Cfrac%7Bp%20-%20P%27%7D%7B%5Csqrt%7B%5Cfrac%7BP%27q%27%7D%7Bn%7D%7D%7D%5D%5Cleq%20%5B%5Cfrac%7B0.634%20-%200.74%7D%7B%5Csqrt%7B%5Cfrac%7B0.74%5Ctimes%200.26%7D%7B60%7D%7D%7D%5D)
P![(p\leq 0.634) = P[z\leq -1.87188]](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87188%5D)
P![(p\leq 0.634) = P[z\leq -1.87] = 0.0298](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87%5D%20%3D%200.0298)
Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %
Thus the on-time rate of 74% is incorrect.
Once the cake is removed from the oven, it has been exposed to the room temperature of 70°F. After 10 minutes, the cake's temperature decreased to 200°F, which is 150°F cooler than it initially was (350°F). As for the what is asked, 90°F is 260°F cooler than its original temperature (350°F).
This problem can be expressed in a ratio:
10mins:150<span>°F = N:260</span><span>°F
where N is how long it will take for</span><span> cake to cool to 90°F
</span>150<span>°FxN=10x260
</span>150N=2600
N=2600/150
N=17.33
N<span>≈20 minutes
Thus, it takes approximately 20 minutes </span><span>for the cake to cool to 90°F</span>
