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2 years ago

What is the probability that a randomly selected date in the year 2021 is not in the month of October or

Mathematics

1 answer:

myrzilka [38]2 years ago

5 0

Answer:

0.833

Step-by-step explanation:

There are 365 days in 2021

There are 31 days in October

There are 30 days in November

(365-61)/365 = 0.833

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Arada [10]

Well the question is written wrong but I think is 30

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2 years ago

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Balancing equations C4H10O+O2>>CO2+h2O

NemiM [27]

Answer:

C4H10O + O2 = CO2 + H2O - Chemical Equation Balancer.

Step-by-step explanation:

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3 years ago

Use the diagram show to calculate m

solniwko [45]

Answer:

131 degrees

Step-by-step explanation:

Angle PNA is 180-149 which is 31 from supplementary angles.

Angle APN is 162 - 31 which is 131 from exterior angles. We were able to find this since an exterior angle of a triangle is just the sum of the other two angles in the triangle combined.

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3 years ago

Solve 3х - х + 2 = 12. a. -5 b. 4 c. 7 d. 5

DaniilM [7]

Answer:

3x-x+2=12

2x+2=12

2x = 10

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6 0

3 years ago

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If A, B, and C are integers between 1 and 10 (inclusive), how many different combinations of A, B, and C exist such that A

Marianna [84]

$\fontsize{18}{10}{\textup{\textbf{The number of different combinations is 120.}}}$

Step-by-step explanation:

A, B and C are integers between 1 and 10 such that A<B<C.

The value of A can be minimum 1 and maximum 8.

If A = 1, B = 2, then C can be one of 3, 4, 5, 6, 7, 8, 9, 10 (8 options).

If A = 1, B = 3, then C has 7 options (4, 5, 6, 7, 8, 9, 10).

If A = 1, B = 4, then C has 6 options (5, 6, 7, 8, 9, 10).

If A = 1, B = 5, then C has 5 options (6, 7, 8, 10).

If A = 1, B = 6, then C has 4 options (7, 8, 9, 10).

If A = 1, B = 7, then C has 3 options (8, 9, 10).

If A = 1, B = 8, then C has 2 options (9, 10).

If A = 1, B = 9, then C has 1 option (10).

So, if A = 1, then the number of combinations is

$n_1=1+2+3+4+5+6+7+8=\dfrac{8(8+1)}{2}=36.$

Similarly, if A = 2, then the number of combinations is

$n_2=1+2+3+4+5+6+7=\dfrac{7(7+1)}{2}=28.$

If A = 3, then the number of combinations is

$n_3=1+2+3+4+5+6=\dfrac{6(6+1)}{2}=21.$

If A = 4, then the number of combinations is

$n_4=1+2+3+4+5=\dfrac{5(5+1)}{2}=15.$

If A = 5, then the number of combinations is

$n_5=1+2+3+4=\dfrac{4(4+1)}{2}=10.$

If A = 6, then the number of combinations is

$n_6=1+2+3=\dfrac{3(3+1)}{2}=6.$

If A = 7, then the number of combinations is

$n_7=1+2=\dfrac{2(2+1)}{2}=3.$

If A = 3, then the number of combinations is

$n_8=1.$

Therefore, the total number of combinations is

$n\\\\=n_1+n_2+n_3+n_4+n_5+n_6+n_7+n_8\\\\=36+28+21+15+10+6+3+1\\\\=120.$

Thus, the required number of different combinations is 120.

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Question : ow many types of zygotic combinations are possible between a cross AaBBCcDd × AAbbCcDD?

Link : https://brainly.in/question/4909567.

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3 years ago