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Keith_Richards [23]

3 years ago

5

two trains leave the station at the same time, one heading west and the other east. The westbound train travels 10 miles per hou

r slower than the eastbound train. if the two trains are 540 miles apart after 3 hours, what is the rate of the westbound train? do not do any rounding.

1 answer:

Travka [436]3 years ago

6 0

Answer: the rate of the westbound train is 85 mph.

Step-by-step explanation:

Let x represent the rate of the westbound train. The westbound train travels 10 miles per hour slower than the eastbound train. It means that the rate of the eastbound train is (x + 10) mph.

The two trains leave the station at the same time. The total distance covered by both trains after 3 hours is 540 miles.

Distance = speed × time

Distance covered by the westbound train in 3 hours is 3x.

Distance covered by the eastbound train in 3 hours is 3(x + 10)

Therefore,

3x + 3(x + 10) = 540

3x + 3x + 30 = 540

6x = 540 - 30 = 510

x = 510/6 = 85 mph

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Pls answer this with a picture showing the correct box plot

faust18 [17]

Answer/Step-by-step explanation:

To find out the mistake of the student, let's find the min, max, median, Q1 and Q3, which make up the 5 important values that are represented in a box plot.

Given, {2, 3, 5, 6, 10, 14, 15},

Minimum value = 2

Median = middle data point = 6

Q1 = 3 (the middle value of the lower part of the data set before the median)

Q3 = 14 (middle value of the upper part of the data set after the median)

Maximum value = 15

If we examine the diagram the student created, you will observe that he plotted the median wrongly. The median, which is represented by the vertical line that divides the box, ought to be at 6 NOT 10.

See the attachment below for the correct box plot.

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3 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

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Answer:

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