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il63 [147K]
2 years ago
10

Two trains left the station traveling in opposite directions. The distance Train A traveled is modeled by the function d(t)=81t

where d represents distance in miles and t represents time in hours.
Train B traveled a total of 324 miles in 4 hours.

How does the distance Train A traveled in 1 hour compare to the distance Train B traveled in 1 hour?



The distance Train A traveled in 1 h is equal to the distance Train B traveled in 1 h.

The distance Train A traveled in 1 h is greater than the distance Train B traveled in 1 h.

The distance Train A traveled in 1 h is less than the distance Train B traveled in 1 h.
Mathematics
1 answer:
stira [4]2 years ago
8 0

Answer:

Option A -  The distance Train A traveled in 1 h is equal to the distance Train B traveled in 1 h.

Step-by-step explanation:

Given : The distance Train A traveled is modeled by the function d(t)=81t

where d represents distance in miles and t represents time in hours.

To find : How does the distance Train A traveled in 1 hour compare to the distance Train B traveled in 1 hour?

Solution :

Distance traveled by Train A in 1 hour is

d(t)=81t

d(t)=81(1)=81

Distance traveled by Train B in 1 hour is

d(t)=81t

d(t)=81(1)=81

or for B, we have 324 miles in 4 hours. If that is at a constant speed, it travels 324/4 = 81 miles in one hour

Therefore, The distance Train A traveled in 1 h is equal to the distance Train B traveled in 1 h.

Hence, Option A is correct.


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The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili
Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that \mu = 74, \sigma = 15

Question a:

Sample of 36 means that n = 36, s = \frac{15}{\sqrt{36}} = 2.5

This probability is 1 subtracted by the pvalue of Z when X = 78. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{78 - 74}{2.5}

Z = 1.6

Z = 1.6 has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that n = 150, s = \frac{15}{\sqrt{150}} = 1.2247

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

Z = \frac{X - \mu}{s}

Z = \frac{77 - 74}{1.2274}

Z = 2.45

Z = 2.45 has a pvalue of 0.9929

X = 71

Z = \frac{X - \mu}{s}

Z = \frac{71 - 74}{1.2274}

Z = -2.45

Z = -2.45 has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that n = 219, s = \frac{15}{\sqrt{219}} = 1.0136

This probability is the pvalue of Z when X = 74.2. So

Z = \frac{X - \mu}{s}

Z = \frac{74.2 - 74}{1.0136}

Z = 0.2

Z = 0.2 has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0
3 years ago
Write the quadratic equation in general form. (x - 3)^ 2 = 0
algol [13]
The answer is:  x² – 6x + 9 = 0  .
_____________________________________________________
Explanation:
________________________________________________________

Given:  (x – 3)² = 0 ;  write as: general form:  "ax² + bx + c = 0";  a ≠ 0 .
<span>
Note:  </span>(x – 3)² = (x – 3)(x – 3) = x² – 3x – 3x + 9 = x² – 6x + 9 ;
___________________________________________________
Rewrite: (x – 3)²  =  0 ;  →
           as:   x² – 6x + 9 = 0  ; which is our answer.
____________________________________________
→   x² – 6x + 9 = 0  ; is in "general form", or "standard equation format"; that is: " ax² + bx + c = 0 ";  (a ≠ 0) ;
             → in which: 
   a =  1 (implied coefficient, since anything multiplied by "1" is that same                                                                                                                    value);
   b = -6; 
   c =  9
_______________________________________________________
5 0
3 years ago
Read 2 more answers
I am an even number i am less than 100,the sum of my digits is 15, i am multiple of 12
Yakvenalex [24]

Answer:

96

Step-by-step explanation:

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7 0
3 years ago
What form is this equation written in? y-7=2(x-6)
Alexxx [7]

Answer:

this is written in point slope form

Step-by-step explanation:

y-y1 = m(x-x1)

the slope is 2

and the point is (6,7)

7 0
3 years ago
If x = 4, which inequality is true?<br> A) -2 -3x 12<br> C) 2x -4 -13<br> Show your work
wolverine [178]

Answer:

true

Step-by-step A

4 0
3 years ago
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