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lozanna [386]
3 years ago
9

What the sum of 57.03 + 2.08​

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
6 0

Answer:

59.11

Step-by-step explanation:

57.03 + 2.08=

59.11

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Answer the problem and show all work.
marishachu [46]

Member Price = 30 + 3x

Non-member Price = 6x

x = the tickets they buy.

We want 30 + 3x = 6x

So, first we subtract 3x from each side and are left with:

30 = 3x

Then, divide each side by 3.

10 = x

So, the cost of 10 tickets is the same more non-members and members.

We can also check it:

Member Price: 30 + 3(10)

30 + 30 = 60

Non-member Price: 6(10)

60

8 0
3 years ago
An instructor gives her class the choice to do 7 questions out of the 10 on an exam.
Maksim231197 [3]

Answer:

(a) 120 choices

(b) 110 choices

Step-by-step explanation:

The number of ways in which we can select k element from a group n elements is given by:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways in which a student can select the 7 questions from the 10 questions is calculated as:

10C7=\frac{10!}{7!(10-7)!}=120

Then each student have 120 possible choices.

On the other hand, if a student must answer at least 3 of the first 5 questions, we have the following cases:

1. A student select 3 questions from the first 5 questions and 4 questions from the last 5 questions. It means that the number of choices is given by:

(5C3)(5C4)=\frac{5!}{3!(5-3)!}*\frac{5!}{4!(5-4)!}=50

2. A student select 4 questions from the first 5 questions and 3 questions from the last 5 questions. It means that the number of choices is given by:

(5C4)(5C3)=\frac{5!}{4!(5-4)!}*\frac{5!}{3!(5-3)!}=50

3. A student select 5 questions from the first 5 questions and 2 questions from the last 5 questions. It means that the number of choices is given by:

(5C5)(5C2)=\frac{5!}{5!(5-5)!}*\frac{5!}{2!(5-2)!}=10

So, if a student must answer at least 3 of the first 5 questions, he/she have 110 choices. It is calculated as:

50 + 50 + 10 = 110

6 0
3 years ago
CAN SOMEONE PLZ ANSWER MY LATEST QUESTION (I give out brainliest)
sveticcg [70]

Answer:

ok can you give the link

Step-by-step explanation:

4 0
3 years ago
In 2012 your car was worth $10,000. In 2014 your car was worth $8,800. Suppose the value of your car decreased at a constant rat
Sonbull [250]

Answer:

g(t) = 10000(0.938)^t

Step-by-step explanation:

Given data:

car worth is $10,000 in 2012

car worth is $8000 in 2014

let linear function is given as

P(t) = at + b

which denote the value of car in year t

take t =0 for year 2012

at t =0, 10,000 = 0 + b

we get b = 10,000

take t =2 for year 2014

at t =2, P(2) = 2a + b

8800 = 2a + 10,000

a = - 600

Thus the price of car at year t after 2012 is given as p(t) = -600t + 10000

let the exponential function p(t)  =ab^2 where t denote t = 0 at 2012

putting t = 0 P(0) = 10,000 we get 10,000 = ab^0

a = 10,000

putting t = 2 p = 8800

8800 =ab^2

b^2 = \frac{8800}{10000}

b = 0.938

g(t) = 10000(0.938)^t

3 0
3 years ago
Simplify the expression so there is only one positive power for each base. (5^-2 x 4^-4)^-2
Oliga [24]
40960000 is the answer to the equation
6 0
3 years ago
Read 2 more answers
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