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ivolga24 [154]
3 years ago
6

A 13 L solution that was 12% vinegar was mixed with a 9 L solution that was 70% vinegar. What is the concentration of the new mi

xture?
Mathematics
1 answer:
erastovalidia [21]3 years ago
6 0

Answer:

The concentration is simply 36%

Step-by-step explanation:

In this question, we are concerned with calculating the concentration of a new mixture formed from mixing some liters of each of two vinegar variants of different concentrations.

We proceed as follows;

The concentration of the new solution will contain 12% of 13L vinegar A and 70% of 9L vinegar B

13L of vinegar A will contain 13 * 12% = 13 * 0.12 = 1.56

9L of 70% vinegar B will contain 9 * 70% = 9 * 0.7 = 6.3

Now, the new mixture has a total volume of 13 + 9 = 22L

The concentration of the new mixture will thus be;

(1.56 + 6.3)/22

= 0.357 and that’s approximately 0.36 or simply 36%

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A sum of money was to be shared among three friends,Albert,Micheal,Moses in the ratio 3:5:6.if Micheal receives $45,find the sum
kirza4 [7]

Answer:

The sum of the money shared is $1372

Step-by-step explanation:

The given ratio is

Albert:Michael:Moses=3:5:6

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Let the sum of money shared be $x.

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Psychologist Michael Cunningham conducted a survey of university women to see whether, upon graduation, they would prefer to mar
bagirrra123 [75]

Answer:

A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.

B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.

C. The expected value of X is 6.75, and the standard deviation of X is 2.17.

Step-by-step explanation:

The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.

With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.25^{k} 0.75^{25-k}\\\\\\

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P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\

B. P(x≥10)

P(x\geq10)=1-P(x

P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.25*0.001=0.0063\\\\\\P(x=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.0625*0.0013=0.0251\\\\\\P(x=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.0156*0.0018=0.0641\\\\\\P(x=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.0039*0.0024=0.1175\\\\\\P(x=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.001*0.0032=0.1645\\\\\\P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.0002*0.0042=0.1828\\\\\\

P(x=7) = \dbinom{25}{7} p^{7}(1-p)^{18}=480700*0.000061*0.005638=0.1654\\\\\\P(x=8) = \dbinom{25}{8} p^{8}(1-p)^{17}=1081575*0.000015*0.007517=0.1241\\\\\\P(x=9) = \dbinom{25}{9} p^{9}(1-p)^{16}=2042975*0.000004*0.010023=0.0781\\\\\\

P(x\geq10)=1-(0.0008+0.0063+0.0251+0.0641+0.1175+0.1645+0.1828+0.1654+0.1241+0.0781)\\\\P(x\geq10)=1-0.9287=0.0713

C. The expected value (mean) and standard deviation of this binomial distribution can be calculated as:

E(x)=\mu=n\cdot p=25\cdot 0.25=6.25\\\\\sigma=\sqrt{np(1-p)}=\sqrt{25\cdot 0.25\cdot 0.75}=\sqrt{4.69}\approx2.17

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