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AVprozaik [17]
4 years ago
7

A president and vice president for a sorority are chosen from ten people. How many different president/vice president combinatio

ns are possible?
Mathematics
1 answer:
Arisa [49]4 years ago
4 0

Answer:

There are 90 ways.

Step-by-step explanation:

We have ten people and we have to chose 2 of those people to be the president and vice president. So first we can chose a president, we have 10 choices for president as there are 10 people.

Now we can chose vice president but this time we only have 9 choices because we already chose a president which gives us 9 other people to chose who the vice president is.

Now we need to multiply 10 and 9 together which gives us the total amount of possible combinations for president and vice president if there are 10 people to chose from.

10*9            multiplying

= 90.

Therefore there are 90 ways for a president and a vice president to be chosen for a sorority from ten people.

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The slope f '(x) at each point (x, y) on a curve. y = f (x) is given along with a particular point (a, b) on the curve. Use this
Advocard [28]

Answer:

Step-by-step explanation:

Integrating each term with respect to x, we get:

            x^3          x^2

f(x) = 9--------- + 4------- - 4x + C

               3            2

We are told that if x = 0, f(x) = -7, and so C must equal - 7.  

The solution is

            x^3          x^2

f(x) = 9--------- + 4------- - 4x - 7,   or   f(x) = 3x^3 + 2x^2 - 4x - 7

               3            2

5 0
3 years ago
What is the equation if its Slope= -3/5, y-intercept =2
Oksana_A [137]

Answer:

y = -3/5x +2

Step-by-step explanation:

y = mx + b

where m is the slope and b is the y intercept

y = -3/5x + 2

3 0
3 years ago
Assortative mating is a nonrandom mating pattern where individuals with similar genotypes and/or phenotypes mate with one anothe
scZoUnD [109]

Answer:

a) P(male=blue or female=blue) = 0.71

b) P(female=blue | male=blue) = 0.68

c) P(female=blue | male=brown) = 0.35

d) P(female=blue | male=green) = 0.31

e) We can conclude that the eye colors of male respondents and their partners are not independent.

Step-by-step explanation:

We are given following information about eye colors of 204 Scandinavian men and their female partners.

              Blue    Brown     Green    Total

Blue        78         23            13          114

Brown     19         23            12          54

Green     11           9             16          36

Total      108       55            41          204

a) What is the probability that a randomly chosen male respondent or his partner has blue eyes?

Using the addition rule of probability,

∵ P(A or B) = P(A) + P(B) - P(A and B)

For the given case,

P(male=blue or female=blue) = P(male=blue) + P(female=blue) - P(male=blue and female=blue)

P(male=blue or female=blue) = 114/204 + 108/204 − 78/204

P(male=blue or female=blue) = 0.71

b) What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=blue) = 78/114

P(female=blue | male=blue) = 0.68

c) What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=brown) = 19/54

P(female=blue | male=brown) = 0.35

d) What is the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=green) = 11/36

P(female=blue | male=green) = 0.31

e) Does it appear that the eye colors of male respondents and their partners are independent? Explain

If the following relation holds true then we can conclude that the eye colors of male respondents and their partners are independent.

∵ P(B | A) = P(B)

P(female=blue | male=brown) = P(female=blue)

or alternatively, you can also test

P(female=blue | male=green) = P(female=blue)

P(female=blue | male=blue) = P(female=blue)

But

P(female=blue | male=brown) ≠ P(female=blue)

19/54 ≠ 108/204

0.35 ≠ 0.53

Therefore, we can conclude that the eye colors of male respondents and their partners are not independent.

7 0
3 years ago
A consumer organization estimates that over a​ 1-year period 20​% of cars will need to be repaired​ once, 5​% will need repairs​
gtnhenbr [62]

Answer:

a) 0.5476

b) 0.0676

c) 0.4524

Step-by-step explanation:

Given this information, we can conclude that 74% of the cars won't need any repairs over a 1-year period (100 - 20 - 5 - 1 = 74%). And 26% will need at least 1 repair over a 1-year period.

P(car doesn't need repair) = 0.74

P (car needs repair) = 0.26

If you own two cars, the probability that:

<u>a) Neither will need repair:</u>

We need that car 1 won't need repair AND car 2 won't need repair.

=P(Car 1 doesn't need repair) x P(Car 2 doesn't need repair)

= 0.74 x 0.74 = 0.5476

The probability that neither will need repair is 0.5476.

<u>b) Both will need repair:</u>

We need that car 1 needs repair AND car 2 needs repair.

P(Car 1 needs repair) x P(Car 2 needs repair)

= 0.26 x 0.26 = 0.0676

The probability that both will need repair is 0.0676

<u>c) At least one car will need repair</u>

Car 1 needs repair or Car 2 needs repair or both need repair.

To solve this one, it's easier to use the complement of P(neither needs repair)

1 - P(neither needs repair)

1 - (0.74)(0.74)  = 1 - 0.5476 = 0.4524

The probability that at least one car will need repair is 0.4524

4 0
3 years ago
In 2004, Cindy had $4000 in a mutual fund account. In 2005, the
gtnhenbr [62]
6,250
1,000/4,000 = 25/100
25/100 = x/5,000
x = 1,250

5,000 + 1,250 = 6,250
4 0
3 years ago
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