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Mazyrski [523]
3 years ago
11

Compute the probability of event E if the odds in favor of E are (A) StartFraction 20 Over 27 EndFraction (B) StartFraction 25 O

ver 18 EndFraction(C) StartFraction 19 Over 15 EndFraction (D) StartFraction 21 Over 8 EndFraction
Mathematics
1 answer:
vovikov84 [41]3 years ago
4 0

Answer:

A) 0.4255

B) 0.5813

C) 0.5588

D) 0.7241

Step-by-step explanation:

Assuming that the probability for each case is the same, we can calculate the  probability by dividing the number of favourable cases with the total amount of cases. You can find the total amount of cases by adding the favourable ones with the non favourable ones, so, the probability of each case would be:

A) 20/(27+20) = 20/47 = 0.4255

B) 25/(18+25) =  25/43 = 0.5813

C) 19/(15+19) = 19/34 = 0.5588

D) 21((8+21) = 21/29 = 0.7241

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Ber [7]
It’s going to be A. She was supposed to use distributive property
4 0
3 years ago
Read 2 more answers
What does 3(x+6)-2 equal
NeTakaya

in simplist form it should be 3x+16

8 0
3 years ago
If you help me with this fast I will brain you.<br> Thanks.
irakobra [83]

Answer:

-2 and -6 is the correct answer

5 0
3 years ago
A muffin recipe, which yields 12 muffins, calls for 2/3 cup of milk for every 1 3/4 cups of flour. The same recipe calls for 1/4
djyliett [7]

Answer:

4\dfrac{3}{8} cups of flour

Step-by-step explanation:

A muffin recipe, which yields 12 muffins, calls for 2/3 cup of milk for every 1 3/4 cups of flour.

Then this recipe, which yields one muffin, calls for

\dfrac{2}{3}:12=\dfrac{2}{3}\cdot \dfrac{1}{12}=\dfrac{1}{18}

cup of milk for every

1\dfrac{3}{4}:12=\dfrac{7}{4}\cdot \dfrac{1}{12}=\dfrac{7}{48}

cups of flour.

Thus,

this recipe, which yields a batch of 30 muffins, calls for

\dfrac{1}{18}\cdot 30=\dfrac{5}{3}=1\dfrac{2}{3}

cups of milk for every

\dfrac{7}{48}\cdot 30=\dfrac{210}{48}=\dfrac{35}{8}=4\dfrac{3}{8}

cups of flour.

4 0
3 years ago
Anonymous 3 years ago
amm1812
Two angles are said to be complementary, if the sum of the measures of the to angles is equal to 90 degrees.

Thus, given that HFG is complementary to ACB, them mHFG + mACB = 90 degrees.

From the figure, given that the line from point F meats line CE at point P, then HFG = CFP.
But mCFE = 90 degrees and mCFE = mCFP + mPFE
Also PFE = DFH

Thus, mCFE = mCFP + mPFE = mHFG + mDFH = 90 degrees

Recall that mHFG + mACB = 90 degrees

Thus, mHFG + mACB = mHFG + mDFH

Therefore, mACB = mDFH.
5 0
3 years ago
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