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2 years ago

Help me out to get the answer please

Mathematics

2 answers:

kogti [31]2 years ago

6 0

Answer:

Step-by-step explanation:

1+6=7

7+4=11

11+11=21

fenix001 [56]2 years ago

5 0

Answer:

m=7

Step-by-step explanation:

it's in the picture

the left got cut out but there's a bracket at the end of the first line

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A limited edition poster increases in value each year. After 1 year, the poster is worth $20.70. After 2 years it is worth $23.8

MA_775_DIABLO [31]

20.70-23.80=x+y= year three and x equals how much they make per a year...

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3 years ago

What is the solution to this equation? 7-1^3 √2-x=12

nydimaria [60]

What is the solution to this equation? 7-1^3 √2-x=12

X=5

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3 years ago

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43 milk jugs are in a fridge with 11 that are spoiled. 5 jugs are randomly selected. what are the odds that the first two are go

malfutka [58]

Answer:

0.85% probability that the first two are good and the last three are spoiled

Step-by-step explanation:

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

The order in which the first two jugs is not important, as is not the order in which the last three are selected. 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.

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

Desired outcomes:

2 spoiled, from a set of 43-11 = 32.

3 spoiled, from a set of 11.

$D = C_{32,2}*C_{11,3} = \frac{32!}{2!(32-2)!}*\frac{11!}{3!(11-3)!} = 982080$

Total outcomes:

5 jugs from a set of 43.

$T = C_{43,5} = \frac{43!}{5!(43-5)!} = 115511760$

Probability:

$p = \frac{D}{T} = \frac{982080}{115511760} = 0.0085$

0.85% probability that the first two are good and the last three are spoiled

4 0

3 years ago

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Ymorist [56]

Answer:

Step-by-step explanation:

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3 years ago

The figures below are similar. What are a) the ratio of the perimeters and b) the ratio of the areas of the larger figure to the

iren [92.7K]

A) ratio of the perimeters is the same as ratio of corresponding sides, so:

$\dfrac{P_1}{P_2}=k=\dfrac{22}{14}=\dfrac{11}{7}\approx1,57$

b) ratio of the areas is equal to k² so:

$\dfrac{A_1}{A_2}=k^2=\left(\dfrac{11}{7}\right)^2=\dfrac{121}{49}\approx2,47$

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3 years ago

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