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RSB [31]
3 years ago
9

Answers for these im strugglinh

Mathematics
1 answer:
Rainbow [258]3 years ago
5 0

Answer:

Step-by-step explanation:

40.

Multiples of 3: 3,6,9,12, 15, …

Multiples of 5: 5, 10, 15, 20, …

LCM of 3 and 5 is 15

Multiples of 6: 6, 12, 18, 24,…

Multiples of 9: 9, 18, 27,…

LCM of 6 and 9 is 18

42.

1/2 and 2/3

denominator 2 has multiples: 2, 4, 6, 8, 10, 12 …

denominator 3 has multiples: 3, 6, 9, 12,…

LCD  of 1/2 and 2/3 is 6

equivalent fraction 1/2 = 3(1/2) = 3/6

equivalent fraction 2/3 = 2(2/3) = 4/6

43.

5/12 and 2/15

denominator 12 has multiples: 12, 24, 36, 48, 60…

denominator 15 has multiples: 15, 30, 45, 60…

LCD  of 5/12 and 2/15 is 60

equivalent fraction 5/12 = 5(5/12) = 25/60

equivalent fraction 2/15 = 4(2/15) = 8/60

44.

16/48 has the prime factorization 2*2*2*2/ 2*2*2*2*3

16/48 simplification is 1/3

45.

12/40 has the prime factorization 2*2*3/2*2*2*5

12/40 simplification is 3/ 10

46.

18/63 has the prime factorization 2*3*3/ 3*3*7

18/63 simplification is 2/7

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1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the
Musya8 [376]

Answer:

A) 34.13%

B)  15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

z=\frac{x-m}{s}

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

                                                =  0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587

C) Between 80 and 120

P( 80

D) less than 80

P( x < 80 ) = P( z

E) Between 70 and 100

P( 70

F) More than 130

P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013

8 0
3 years ago
two nascar cars are racing around a 2 mile track. one car can complete a lap in 48 seconds, and the other car can complete one l
Citrus2011 [14]
Let's assume that the faster car has a 2 mile "head start".
The faster car is 1.2 times faster (48/40) than the slower car.
It will take five 2 mile trips to "catch" the other car.
Five 2 mile trips will take 5 * 40 seconds = 200 seconds.



6 0
3 years ago
Read 2 more answers
Find the value of 13 + (-7)
Vikki [24]

Answer:

6

Step-by-step explanation:

Before solve this you've to know that,

( + ) × ( - ) = ( - )

( + ) × ( + ) = ( + )

( - ) × ( - ) = ( + )

Let us solve now.

13 + (-7)

13 - 7

= 6

Hope this helps you :-)

Let me know if you have any other questions  :-)

7 0
2 years ago
Read 2 more answers
(1 point)
aniked [119]

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Point W ____ is on segment AB so that AW= 1/5AB
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The answer is 1.3.4
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