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Luda [366]
3 years ago
8

Solve 3[-x + (2 x + 1)] = x - 1.

Mathematics
1 answer:
vivado [14]3 years ago
8 0

Answer:

The answer is-2.

Step-by-step explanation:

#Hope it helps uh......

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The sum of two numbers is 48. If one third of one number is 5 greater than one sixth of another number, which of the following i
lubasha [3.4K]
<span>The sum of two numbers is 48.
a + b = 48
;
If one third of one number is 5 greater than one sixth of another number,
a = b + 5
multiply both sides by 6, cancel the fractions
2a = b + 30
2a - b = 30
</span><span>use elimination to solve this
a + b = 48
2a - b =30
-------------Addition eliminates b, find a
3a = 78
a = 
a = 26
then
26 + b = 48
b = 48 - 26

b = 22</span>
5 0
3 years ago
Cheryl's making trail mix for a friend she always uses three cups of almonds for every four cups of cashews if Cheryl wants to m
Mkey [24]

Answer:

c

Step-by-step explanation:

3x3=9

4x3=12

4 0
3 years ago
Read 2 more answers
All employees work less than 40 hours
bezimeni [28]
Nice..................
3 0
3 years ago
of 320 consumers polled, some purchase blu-ray movies, some only purchase movies on dvd, and some do both. if 300 people polled
Helga [31]

Answer:

20 people only buy dvds.

Step-by-step explanation:

There are two groups in this problem. One group of people that buys dvds and one that buys blu rays. The total amount of people who buy blu rays is 300, but there's an intersection between the groups and the size of this is 280 people who actually buy both. In order to find out how many people only buy DVDs we first need to figure out how many only buy blu-rays. That is:

people who only buy blu rays = people who buy blu rays - people who buys blu rays and dvds

people who only buy blu rays = 300 - 280 = 20

We can now calculate the amount of people who only buy dvds and that is:

people who only buy dvds =total amount of people - ( people who only buy blu rays + people who buy both)

people who only buy dvds = 320 - (20 + 280) = 320 - 300 = 20 people

3 0
4 years ago
In a certain Algebra 2 class of 30 students, 14 of them play basketball and 10 of them play baseball. There are 14 students who
Veseljchak [2.6K]

Answer:

In a certain Algebra 2 class of 30 students, 22 of them play basketball and 18 of them play baseball. There are 3 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

I know how to calculate the probability of students play both basketball and baseball which is 1330 because 22+18+3=43 and 43−30 will give you the number of students plays both sports.

But how would you find the probability using the formula P(A∩B)=P(A)×p(B)?

Thank you for all of the help.

That formula only works if events A (play basketball) and B (play baseball) are independent, but they are not in this case, since out of the 18 players that play baseball, 13 play basketball, and hence P(A|B)=1318<2230=P(A) (in other words: one who plays basketball is less likely to play basketball as well in comparison to someone who does not play baseball, i.e. playing baseball and playing basketball are negatively (or inversely) correlated)

So: the two events are not independent, and so that formula doesn't work.

Fortunately, a formula that does work (always!) is:

P(A∪B)=P(A)+P(B)−P(A∩B)

Hence:

P(A∩B)=P(A)+P(B)−P(A∪B)=2230+1830−2730=1330

7 0
3 years ago
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