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2 years ago

9

Mrs. Wilson packed supply bags for her students to use in a science experiment. In each bag she put 3 rubber bands and 4 paper c

lips. If she used 48 rubber bands, how many paper clips did she use?

1 answer:

n200080 [17]2 years ago

3 0

Answer:

64

Step-by-step explanation:

In each supply bag, Mrs. Wilson packed 3 rubber bands and 4 paper clips for her students.

Now, it is given that she had used 48 rubber bands.

Hence, there will be (48/3) =16 supply bags that had been used.

Now, in 16 supply bags, there will be (16×4) =64 paper clips.

Therefore, 64 paper clips that she had used. (Answer)

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(-7, -8), (8, -2) distance

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Answer:

3√29

Step-by-step explanation:

Use the distance formula to determine the distance between the two points.

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

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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:

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

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In point f:

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

$= 0.3$

In point g:

$\to (A \cup B) = \frac{7}{10}$

$=0.7$

In point h:

$\to p(A \cap B) = 0.3 \neq 0$

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