The accompanying table shows the numbers of male and female students in a particular

3 answers
Question:

The accompanying table shows the numbers of male and female students in a particular country who received​ bachelor's degrees in business in a recent year. complete parts​ (a) and​ (b) below. business degrees nonbusiness degrees total male 189131 634650 823781 female 169539 885329 1054868 total 358670 1519979 1878649 (a) find the probability that a randomly selected student is male​, given that the student received a business degree. the probability that a randomly selected student is​ male, given that the student received a business​ degree, is nothing. ​(round to three decimal places as​ needed.) ​(b) find the probability that a randomly selected student received a business​ degree, given that the student is female. the probability that a randomly selected student received a business​ degree, given that the student is​ female, is nothing. ​(round to three decimal places as​ needed.)

Answers

Solution:

Probability of an event [tex]=\frac{\text{Total favorable outcome}}{\text{Total possible outcome}}[/tex]

The table showing numbers of male and female students in a particular country who received​ bachelor's degrees in business in a recent year.

         business degrees   Non-business degrees         total

male           189131                  634650                            823781

female       169539                885329                             1054868

total           358670                 1519979                            1878649

We will use Bay's theorem to answer this question.

M=Male, F= Female, B=Business degree

(a)Probability that a randomly selected student is male​, given that the student received a business degree.

[tex]P(\frac{M}{B})=\frac{P(\frac{B}{M})}{P(\frac{B}{M})+P(\frac{B}{F})}=\frac{\frac{189131}{358670}}{\frac{189131}{358670}+\frac{169539}{358670}}\\\\ =\frac{189131}{189131 + 169539}\\\\ =\frac{189131}{358670}\\\\ =0.527[/tex]

You can calculate it directly,

[tex]P(\frac{M}{B})=\frac{\text{Total number of males having business degree}}{\text{Total number of business degree students}}=\frac{189131}{358670}=0.527[/tex]

(b)  The probability that a randomly selected student received a business​ degree, given that the student is Female

[tex]P(\frac{B}{F})=\frac{\text{Girls having business degree}}{\text{Total number of girls}}\\\\ P(\frac{B}{F})=\frac{169539}{1054868}\\\\ P(\frac{B}{F})=0.161[/tex]

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube

Recall that area of the triangle is 1/2*(base)*(height)1/2*(13)*(17) = 110.5the answer is 110.5

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