Answer:
You can see that the chance of having at least one boy is 1/2 in the first generation, 3/4 in the second, and 7/8 in the third.
Step-by-step explanation:
Check the picture below to the left, let's use those sides with the law of sines
![\textit{Law of sines} \\\\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{sin(14^o)}{97}=\cfrac{sin(84^o)}{XZ}\implies XZ = \cfrac{97\cdot sin(84^o)}{sin(14^o)}\implies XZ \approx 398.76 \\\\\\ \stackrel{\textit{now using SOH CAH TOA}}{cos(82^o) = \cfrac{XW}{XZ}}\implies XZcos(82^o)=XW \\\\\\ 398.76cos(82^o)\approx XW\implies 55.497\approx XW\implies \stackrel{\textit{rounded up}}{55=XW}](https://tex.z-dn.net/?f=%5Ctextit%7BLaw%20of%20sines%7D%20%5C%5C%5C%5C%20%5Ccfrac%7Bsin%28%5Cmeasuredangle%20A%29%7D%7Ba%7D%3D%5Ccfrac%7Bsin%28%5Cmeasuredangle%20B%29%7D%7Bb%7D%3D%5Ccfrac%7Bsin%28%5Cmeasuredangle%20C%29%7D%7Bc%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7Bsin%2814%5Eo%29%7D%7B97%7D%3D%5Ccfrac%7Bsin%2884%5Eo%29%7D%7BXZ%7D%5Cimplies%20XZ%20%3D%20%5Ccfrac%7B97%5Ccdot%20sin%2884%5Eo%29%7D%7Bsin%2814%5Eo%29%7D%5Cimplies%20XZ%20%5Capprox%20398.76%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bnow%20using%20SOH%20CAH%20TOA%7D%7D%7Bcos%2882%5Eo%29%20%3D%20%5Ccfrac%7BXW%7D%7BXZ%7D%7D%5Cimplies%20XZcos%2882%5Eo%29%3DXW%20%5C%5C%5C%5C%5C%5C%20398.76cos%2882%5Eo%29%5Capprox%20XW%5Cimplies%2055.497%5Capprox%20XW%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B55%3DXW%7D)
Answer:
The answer is option D
g(x) = |x+3|
Step-by-step explanation:
Please see attached graph
The changes made to the function result from translating the graph of f(x) three units to the left
By substituting the equations provided at x = -3 and x = 0 respectively, we can find the same value for both functions.
g(x = -3) = |(-3)+3| = |0| = 0
f(x = 0) = |0| = 0
Answer:
The fourth term in the sequence is equal to the third term plus the common difference, or 16+3=19- last choice
<span>half-life means you divide by 2
21.6 / 3.6 = 6 hf (hf = half-life)
then you divide by 2^6
6.02 x 10^23 / 2^6 = 9.40625 x 10^21 atoms
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26
3hf ===> 2^3
10.0 / 2^3 = 1.25 gram (remaining)
answer : 10 - 1.25 = 8.75g
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27
128 / 2^n = 2
64 = 2^n
n = 6
24 /6 = 4
4 days is the half-live of the sample
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28
2.5/4.46 of 50% initial mass
(2.5/4.46) x 0.5 x 2 = 0.5605 g
answer: 2 - 0.5605 = 1.4395 g
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29
2^7 = 128
so 7 half-lifes
7* 8040 = 56200 days
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30
the time of 10 half-lifes
= 10 x 0.334s = 3.34s </span><span>
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