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cluponka [151]
3 years ago
12

A freight train leaves a station at noon traveling at 30 mph. A passenger train leaves 1 hour later traveling at 50 mph. At what

time will the passenger train overtake the freight train?
Mathematics
2 answers:
miv72 [106K]3 years ago
8 0
30t=50(t-1)
30t=50t-50
30t-50t=-50
-20t=-50
t=-50/-20
t=2.5 hours after the freight train leaves the station they will meet

zysi [14]3 years ago
4 0

Answer:

after 2.5 hour passenger train would overtake .

Step-by-step explanation:

A freight  train leaves a station at noon traveling at 30 mph.

A passenger train leaves 1 hour later travelling at 50 mph

Let the time to leave the station of freight train = t

time to leave passenger train = t - 1

Since distance = speed × time

the expression will be :

30t = 50(t - 1)

30t = 50t - 50

30t - 50t = -50

-20t = -50

t = \frac{-50}{-20}

t = 2.5 hours

After 2.5 hours passenger train would overtake the freight train.

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The x intercepts occur when you set the factors equal to 0. x+1=0 & x-6=0 gives you the x-intercepts. x+1=0 -> x=-1. x-6=0 -> x=6. Therefore, -1 & 6 are your x intercepts
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3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! I CANNOT RETAKE THIS!!<br><br> Simplify.
NNADVOKAT [17]

Answer:  \frac{2x(5x + 1)}{4x - 1}

<u>Step-by-step explanation:</u>

\frac{4x^{3}-12x^{2}}{4x^{2}+7x - 2} ÷ \frac{2x^{2}-6x}{5x^{2}+11x + 2}

= \frac{4x^{2}(x - 3)}{(4x - 1)(x + 2)} ÷ \frac{2x(x - 3)}{(5x + 1)(x + 2)}

= \frac{4x^{2}(x - 3)}{(4x - 1)(x + 2)} x \frac{(5x + 1)(x + 2)}{2x(x - 3)}

= \frac{2x(5x + 1)}{4x - 1}

8 0
3 years ago
Which situation could be represented by the graph?
Ymorist [56]

The correct option which resembles with the given graph is \fbox{\begin\\\ optionA and optionC\\\end{minispace}}.

Further explanation:

From the given graph the it is observed that the interval is [8,11) which is a half open interval or it can be said that it is a half closed interval.

A half closed or a half open interval is an interval in which one end point of the interval is included but the other point is not included.

For example:

In an interval of the form [a,b) the point a is included in the set but the point b is not included in the set.

Similarly, in the interval [8,11) the point 8 is included in the set but the point 11 is not included in the set.

In order to determine the correct option which resembles with given graph form interval for each option.

Option A:

As per the statement in option A the age of the student must be at least 8 years and not more than 11 years that is the age of the student is either 8 years or greater than 8 years and less than 11 years.

As per the above statement the interval formed is \fbox{\begin\\\ [8,11)\\\end{minispace}} in which 8 is included in the set and 11 is not included which means it is a half closed or half opoen interval.

Therefore, \fbox{\begin\\\ optionA\\\end{minispace}} is a correct option.

Option B:

As per the statement in option B the amount of money the babysitter can earn is between \$8 and \$11 per hour.

As per the above statement the interval formed is \fbox{\begin\\\ (8,11)\\\end{minispace}} in which both of the end points are not included in the set which means that it is an open interval.

Therefore, \fbox{\begin\\\ optionB\\\end{minispace}} is an incorrect option.

Option C:

As per the statement in option C the time required to complete a program in college cannot be less than 8 hours and the time must be less than 11 hours.

From the above statement it is concluded that the time required can be 8 hours or more than 8 hours but it has to be less than 11 hours.

So, as per the above the statement the interval formed is \fbox{\begin\\\ [8,11)\\\end{minispace}}.

Therefore, \fbox{\begin\\\ optionC\\\end{minispace}} is a correct option.

Option D:

As per the statement in option D the age of the dogs is either 8 years old or below 8 years and age of the dog must be greater than  11 tears.

As per the above statement the interval formed is \fbox{\begin\\\ (-\infty,8) \bigcup(11,\infty)\\\end{minispace}}.

Therefore, the \fbox{\begin\\\ optionD\\\end{minispace}} is an incorrect option.

So, \fbox{\begin\\\ optionA and optionC\\\end{minispace}} are the correct options which resembles with the given graph.

Learn more:

  1. A problem on composite function brainly.com/question/2723982
  2. A problem to find radius and center of circle brainly.com/question/9510228
  3. A problem to determine intercepts of a line brainly.com/question/1332667

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Sets

Keywords: Interval, set, open interval, closed interval, half open interval, half closed interval, set theory, program, community college, babysitters, veterinary clinic, ages, students.

5 0
2 years ago
Read 2 more answers
Can someone help me please I need the mode, median, range and the mean.?
Neko [114]

Answer:

First the mode.  Since 5 popped up the most, 5 is the mode.

Next is the median.  I crossed 1 dot from each side until it shows the last dot, and 5 was the last one.

After that the range.  9-2=7

Finally the worst, the mean... 2+2+3+3+3+4+5+5+5+5+5+6+6+6+8+9+9+9+9

=104/19=5.47

SO, Mode=5 Median=5 Range=7, and the mean is 5.47 (rounded nearest hundred)

7 0
3 years ago
PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST
ohaa [14]

Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

7 0
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