can you help me solve the second and third part to this problem?

label Mathematics
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schedule 1 Day
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May 20th, 2015

So the second one is all the animals located in the union of E and B, so 109 + 30 + 19 + 115 + 25 + 28 = 326

The third one is the union of G with the intersection of A and F, so 117 + 57 = 174

Hope this helps let me know if you have any questions

May 20th, 2015

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