A number plus that number squared is equal to 110.

Mathematics

2 answers:

Mamont248 [21]3 years ago

7 0

Answer:

10+10^2

Step-by-step explanation:

If you do 10 and add that squared number of 10, it would be 10+ 100 which is 110.

Art [367]3 years ago

3 0

Answer:

x=10,-11

Step-by-step explanation:

x+x^2=110

10+11^2=110

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You are wearing a pair of cargo pants with six pockets. You've put $10 in one of the pockets, but you cannot remember which one.

Over [174]

There are 6 pockets all together. You checked two of them and it wasn't
there. So you can forget about those 2, and now there are 4 pockets left.

The total number of ways for your next try is 4.
The number that will be successful is 1.

The probability of success on the next one is 1/4 = 25% .

=======================================================

Another method of solution, somewhat more complex:

Total number of pockets = 6.
Amount of money missing = $10
First partial sum, i.e. sum of numbers
in the first part of the problem = 6 + 10 = 16.

Total number of pockets = 6
Current value of cargo pants on a fashion scale from 1 to 10 = -2.
Second partial sum = 6 - 2 = 4.

Probability = (second partial sum) / (first partial sum) = 4/16 = 25% .

5 0

3 years ago

If f(2) =20 and f(3) =10 F(4) = ,f(5)=

Arada [10]

F(4)=0 and F(5)= -10

5 0

3 years ago

Assume that there are an equal number of births in each month so that the probability is that a person chosen at random was born

NikAS [45]

Answer:

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have a birthday in May, or they do not. The probability of a person having a birthday in May is independent of any other person. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$