MATHEMATICS HIGH SCHOOL

PLS HELP! DUE IN 10 MINS PLS!!!!!!!!!!!!!!!

Answers

Answer 1

Answer:

Answer:

25 metres per second

Step-by-step explanation:

just divide 100 by 4 and you will get the answer.☺️☺️

Answer 2

Answer:

Answer:

25 m/s

Step-by-step explanation:

train a travels at 12.5 m/s and will arrive at 100m in 7s.

train b arrives at the same distance, 100m, in 4s. so that means train b is going at 100m/4s and then simplify that

therefore train b is moving at 25m/s

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What is an
equation of the line that passes through the points (0,6) and (-3,2)?

Answers

Answer:

Step-by-step explanation:

y=mx+b is the general equation of a line, where

m= slope = (y2-y1)/(x2-x1)= (6-2)/(0- -3)= 4/3

b= y intercept

for point (0,6)

y=(4/3)x+b will become 6=(4/3)*0 +b, so b=6

the equation of the line that passes trough the given points is

y= (4/3) x +6

Answer:

y = 4/3 x + 6

Step-by-step explanation:

M = -4/-3 = 2/3

y = 4/3 X + B

6 = 4/3(0) + B

B=6

What is the product?(2x – 1)(x + 4)
2x2 – 4
2x2 + 4
2x2 + 7x – 4
2x2 – 7x – 4

Answers

The answer is 2x^2+7x-4 i think that's what you meant for the last one.

Answer:

the answer is 2x^2+7x-4

Step-by-step explanation:

hope it helpsssssssssss

Please answer ASAP! Your answer MUST include an explanation in order to receive points and the Brainliest answer. Thank you.

Answers

63 because if you look at the pattern each number is multiplied by seven so if you multiply the input which is nine by seven you would get 63.

in an ordinary day with no discount if u see the parretn you can get the cost per ticket by dividing the cost by number of tickets
1st : 21/3 =7
2nd: 35/5=7
3rd: 56/8 = 7
so the ticket in an ordinary day = $7
in friday they make discount by $2 so the new value of the ticket = the old value -2
so the ticket on friday = $5
so if you want 9 tickets
9*5 = $45

4(1-3n)-14> 4(2n+3)-9n

Answers

4(1-3n)-14>4(2n+3)-9n
4-12n-14>8n+12-9n
-12n-10>-n+12
-22>11n
-2>n
n<-2

The average reading score on a certain test is given by y=0.150x +255.34 where x is the number of years past 1970 . In what year would the average reading score be 259.24 if this model was accurate

Answers

Answer:

The average was 259.24 in 1996.

Step-by-step explanation:

In order to find the year that the average, "y", is 259.24 we need to apply this value on the given expression, as done below:

$259.24 = 0.15*x + 255.34\n0.15*x = 259.24 - 255.34\n0.15*x =3.9\nx = (3.9)/(0.15)\nx = 26$

The average was 259.24 after 26 years, therefore it was in 1996.

You arrive at a bus stop at 10 a.m., knowing that the bus will arrive at some time uniformly distributed between 10 and 10:30. What is the probability that you will have to wait longer than 10 minutes? If, at 10:15, the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes?

Answers

Answer:

a) the probability of waiting more than 10 min is 2/3 ≈ 66,67%

b) the probability of waiting more than 10 min, knowing that you already waited 15 min is 5/15 ≈ 33,33%

Step-by-step explanation:

to calculate, we will use the uniform distribution function:

p(c≤X≤d)= (d-c)/(B-A) , for A≤x≤B

where p(c≤X≤d) is the probability that the variable is between the values c and d. B is the maximum value possible and A is the minimum value possible.

In our case the random variable X= waiting time for the bus, and therefore

B= 30 min (maximum waiting time, it arrives 10:30 a.m)

A= 0 (minimum waiting time, it arrives 10:00 a.m )

a) the probability that the waiting time is longer than 10 minutes:

c=10 min , d=B=30 min --> waiting time X between 10 and 30 minutes

p(10 min≤X≤30 min) = (30 min - 10 min) / (30 min - 0 min) = 20/30=2/3 ≈ 66,67%

a) the probability that 10 minutes or more are needed to wait starting from 10:15 , is the same that saying that the waiting time is greater than 25 min (X≥25 min) knowing that you have waited 15 min (X≥15 min). This is written as P(X≥25 | X≥15 ). To calculate it the theorem of Bayes is used

P(A | B )= P(A ∩ B ) / P(A) . where P(A | B ) is the probability that A happen , knowing that B already happened. And P(A ∩ B ) is the probability that both A and B happen.

In our case:

P(X≥25 | X≥15 )= P(X≥25 ∩ X≥15 ) / P(X≥15 ) = P(X≥25) / P(X≥15) ,

Note: P(X≥25 ∩ X≥15 )= P(X≥25) because if you wait more than 25 minutes, you are already waiting more than 15 minutes

- P(X≥25) is the probability that waiting time is greater than 25 min

c=25 min , d=B=30 min --> waiting time X between 25 and 30 minutes

p(25 min≤X≤30 min) = (30 min - 25 min) / (30 min - 0 min) = 5/30 ≈ 16,67%

- P(X≥15) is the probability that waiting time is greater than 15 min --> p(15 min≤X≤30 min) = (30 min - 15 min) / (30 min - 0 min) = 15/30

therefore

P(X≥25 | X≥15 )= P(X≥25) / P(X≥15) = (5/30) / (15/30) =5/15=1/3 ≈ 33,33%

Note:

P(X≥25 | X≥15 )≈ 33,33% ≥ P(X≥25) ≈ 16,67% since we know that the bus did not arrive the first 15 minutes and therefore is more likely that the actual waiting time could be in the 25 min - 30 min range (10:25-10:30).