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3 years ago

PLZ HELP PLZ I just need the ratio

Mathematics

2 answers:

PIT_PIT [208]3 years ago

7 0

8:12 and 36:54

Explanation: I copied the person from above

iris [78.8K]3 years ago

5 0

Answer: 8:12 and 36:54

Step-by-step explanation:

You might be interested in

2 to the Third power -(30÷5) +9

OLga [1]

Answer:

Step-by-step explanation:

2^3-(30/5)+9

First, you divide 30/5 because it is in parentheses and we are following this rule for order of operations:

P- parentheses

E- exponents

M- multiplication

D- division

A- addition

S- subtraction

so 30/5 = 6 and plugged in is:

2^3-(6)+9

Next, we do exponents:

2^3 = 2x2x2 = 8

plugged back in:

8-(6)+9

Now, we do the math from left to right so:

8-6=2

2+9=11

11 is your answer

8 0

3 years ago

A rectangles width is 1/4 of its length. It’s area is 9 square units. The equation l(1/4l) =9 can be used to find l, the length

Zielflug [23.3K]

Answer: the answer is 6

Step-by-step explanation: because 6/4 = 1.5 and 1.5• 6 =9

3 0

2 years ago

Five computer program modules are ranked as M1, M2, M3, M4, and M5 according to the ascending order of effort required to debug

gulaghasi [49]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Technician selects three out of 5 systems

In C(5,3)=10ways, this can be achieved

In part a:

Space sample chooses 3 of a 5 systems

$(M_1, \ M_2,\ M_3),$ $(M_1,M_2,M_4) \ (M_1,M_2,M_5) \ (M_1,M_3,M_4) \ (M_1,M_3,M_5),$ $(M_1,M_4,M_5) \ (M_2,M_3,M_4)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5)}$

In point b:

A =MODULE WHICH INCLUDE M1 minimal amount of effort

Outcomes probable =

$(M_1,M_2,M_3),\ (M_1,M_2,M_4) \ (M_1,M_2,M_5)\ (M_1,M_3,M_4)\\\\(M_1,M_3,M_5),\ (M_1,M_4,M)5)\ =\ 6$

$\to p(A)=\frac{6}{10}\\\\$

$=0.6$

In point c:

B = highest effort that is $M_5$

Potential result=

$(M_1,M_2,M_5) \ (M_1,M_3,M_5) \ (M_2,M_3,M_5)\(M_2,M_4,M_5) \\ (M_2,M_4,M_5), \ (M_3,M_4,M_5) \ =\ 6 \\\\$

$\to B= \frac{6}{10} \\\\$

$=0.6$

$\to P(B)=10$

In point d:

$\to \ A \ intersection \ B=(M_1,M_2,M_5), \ (M_1,M_3,M_5) \ ,(M_1,M_4,M_5)$

$\to A (A \ intersection \ B) = \frac{3}{10} \\\\\ \ \ \ \ \$

$=0.3$

In point e:

$\to (A \cup B) = (M_1,M_2,M_3),\ (M_1,M_2,M_4)\ (M_1,M_2,M_5)(M_1,M_3,M_4)\ (M_1,M_3,M_5), \\ (M_1,M_4,M_5)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5) \ = \ 9$ $\to P(A \cap B)=\frac{9}{10}$