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Virty [35]
3 years ago
6

If a couple were planning to have three​ children, the sample space summarizing the gender outcomes would​ be: bbb,​ bbg, bgb,​

bgg, gbb,​ gbg, ggb, ggg.
a. construct a similar sample space for the possible weightweight outcomes​ (using o for overweight and u for underweighto for overweight and u for underweight​) of two children.
b. assuming that the outcomes listed in part​ (a) were equally​ likely, find the probability of getting two underweightunderweight children.
c. find the probability of getting exactly one overweightoverweight
Mathematics
1 answer:
MaRussiya [10]3 years ago
5 0
There are two choices for each child: overweight (o) or underweight (u). So if the first child is o the next can be o or u. If the first is u the second can be o or u. This gives four possibilities. Here the first child is the letter noted first and the second is the one listed second:
OO
OU
UO
UU

There are 4 outcomes and if each is equally likely then the probability of each is 1/4. Thus the probability of UU is 1/4

The probability of one underweight and one over weight is 1/2 because in two of the outcomes listed above there is one O and one U (namely OU and UO). Since there are 4 outcomes the probability is 2/4 = 1/2

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AleksAgata [21]

Answer: Three quarters (30 ml) is made up of a 36% acid solution.

Step-by-step explanation:

Since we have given that

Total amount of solution = 40 ml

One- quarter of the solution is given by

\frac{1}{4}\times 40\ ml\\\\=10\ ml

Three-quarter of the solution is given by

\frac{3}{4}\times 40\\\\=30\ ml

According to question, we have

10 ml of solution is made up of = 20% of acid solution

30 ml of solution is made up of = say<em> 'c </em>' of acid solution

40 ml of solution is made up of = 32 % of acid solution

So, our equation becomes,

10\times 0.2+30\times c=40\times 0.32\\\\2+30c=12.8\\\\30c=12.8-2\\\\30c=10.8\\\\c=\frac{10.8}{30}\\\\c=0.36\\\\c=36\%

Hence,  three quarters (30 ml) is made up of a 36% acid solution.

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3 years ago
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A. \frac{27}{19}

To find greater than or smaller than relation, we multiply the terms like (numerator of L.H.S with denominator of R.H.S and put the value on the left side. Then multiply the denominator of L.H.S with numerator of R.H.S and put the value on right side. Now compare the digits.)

So, solving A,  we get 810<209  ... This is false

B. \frac{17}{31}>\frac{19}{14}

= 238>589   ..... This is false

C. \frac{16}{26}>\frac{30}{31}

= 496>780    .... This is false

D. \frac{35}{30}

= 420<660   ..... This is true

Hence, option D is true.

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