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sammy [17]
3 years ago
9

This id hard please help me do this

Mathematics
2 answers:
svlad2 [7]3 years ago
7 0
In this problem, we're going to use the 'PEMDAS' method.

First, we're doing the problem in the parenthesis.
44 - 4 = 40

Next in PEMDAS, we need to do the exponent.
6² = 6 x 6 = 36

Then we would do multiplication, but there is nothing to multiply so we move onto division.
40 / 2 = 20

Now we would add, but there is nothing to add so we move onto subtraction.
20 - 36 = -16

Our final answer would be -16.

Hope this helps!~
Blizzard [7]3 years ago
4 0
44-4=40.
6^2 =6×5=36
80÷2=20
20-36=16
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A club is choosing 2 members to serve on a committee. The club has nominated 2 women and 4 men. Based on chance alone, what is t
tekilochka [14]

Answer:

53.33% probability that one woman and one man will be chosen to be on the committee

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the members are chosen is not important, so we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

What is the probability that one woman and one man will be chosen to be on the committee?

Desired outcomes:

One woman, from a set of 2, and one man, from a set of 4. So

D = C_{2,1}*C_{4,1} = \frac{2!}{1!1!}*\frac{4!}{1!3!} = 8

Total outcomes:

Two members from a set of 2 + 4 = 6. So

T = C_{6,2} = \frac{6!}{2!4!} = 15

Probability:

p = \frac{D}{T} = \frac{8}{15} = 0.5333

53.33% probability that one woman and one man will be chosen to be on the committee

5 0
3 years ago
The marketing club at school is opening a student store. They randomly survey 50 students about how much money they spend on lun
Luden [163]

Answer:

The expected value for a student to spend on lunch each day = $5.18

Step-by-step explanation:

Given the data:

Number of students______$ spent

2 students______________$10

1 student________________$8

12 students______________$6

23 students______________$5

8 students_______________$4

4 students_______________$3

Sample size, n = 50.

Let's first find the value on each amount spent with the formula:

\frac{num. of students}{sample size} * dollar spent

Therefore,

For $10:

\frac{2}{50} * 10 = 0.4

For $8:

\frac{1}{50} * 8 = 0.16 l

For $6:

\frac{12}{50} * 6 = 1.44

For $5:

\frac{23}{50} * 5 = 2.3

For $4:

\frac{8}{50} * 4 = 0.64

For $3:

\frac{4}{50} * 3 = 0.24

To find the expected value a student spends on lunch each day, let's add all the values together.

Expected value =

$0.4 + $0.16 + 1.44 +$2.3 + $0.64 + $0.24

= $5.18

Therefore, the expected value for a student to spend on lunch each day is $5.18

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