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3 years ago

4. Find the average acceleration of a northbound subway train that slows down from 12 m/s

Mathematics

1 answer:

VikaD [51]3 years ago

5 0

Answer:

-3.125m/s^2

Step-by-step explanation:

Average acceleration = v-u /t

That is initial final velocity v minus initial velocity u divided by the time t In seconds .

From the question,

v =9.5m/s

u =12m/s

t =0.8 secs

Using the above formula,

We have

average a = 9.5 -12 /0.8

= -2.5/0.8

= -3.125m/s^2

Negative answer indicates that the train was decelerating

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You get scores of 85 and 91 on two history test.Wright and solve an inequality to find the scores you can get on your next histo

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Let x = score on next test
(85+91+x)/3 $\geq$ 90
x $\geq$ 94

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3 years ago

Hey could someone please help me with this really hard assignment is taking me forever. Thank you marking brainliest!

rewona [7]

Answer:

Exercise 1: The length of the unknown leg is 4 inches.

Exercise 3: The following straws can be used to construct the triangle: b) $3\,in$ , c) $4.5\,in$ , d) $6.5\,in$ , e) $10\,in$ , f) $13.5\,in$

Exercise 4: Possible options of this exercise: 1) (2, 4, 5), 2) (4, 5, 6), 3) (5, 6, 10), 4) (1, 2, 11), 5) (1, 4, 11), 6) (1, 10, 11)

Step-by-step explanation:

Exercise 1:

Let suppose that triangle represented in the figure is a right triangle, the length of the missing leg is determined by Pythagorean Theorem:

$y = \sqrt{l^{2}-x^{2}}$ (1)

Where:

$l$ - Hypotenuse, in inches.

$x$ - Known leg, in inches.

$y$ - Unknown leg, in inches.

If we know that $l = 11\,in$ and $x = 10\,in$ , then the length of the unknown leg is:

$y = \sqrt{21}$

Since 4 is the least whole number closest to $\sqrt{21}$ , then we conclude that the length of the unknown leg is 4 inches.

Exercise 3:

The range of possible lengths for the missing side of the triangle is represented by the following simultaneous inequality:

$x + y > l > x-y$ (2)

Where:

$x$ - Greater side, in inches.

$y$ - Lesser side, in inches.

$l$ - Missing side, in inches.

If we know that $x = 8\,in$ and $y = 6\,in$ , then we have the following range of missing sides:

$14\,in > l > 2\,in$

The following straws can be used to construct the triangle: b) $3\,in$ , c) $4.5\,in$ , d) $6.5\,in$ , e) $10\,in$ , f) $13.5\,in$

Exercise 4:

Let check each pair to determine possible constructions by means of the inequality used in Exercise 3:

(i) $x = 4\,in, y = 2\,in$

$6\,in>l>2\,in$

Possible choices: 5 inches.

(ii) $x = 5\,in, y = 2\,in$

$7\,in > l > 3\,in$

Possible choices: 4 inches, 6 inches.

(iii) $x = 6\,in, y = 2\,in$

$8\,in > l > 4\,in$

Possible choices: 5 inches, 6 inches.

(iv) $x = 10\,in, y = 2\,in$

$12\,in > l > 8\,in$

Possible choices: None.

(v) $x = 5\,in, y = 4\,in$

$9\,in > l > 1\,in$

Possible choices: 2 inches, 6 inches.

(vi) $x = 6\,in, y = 4\,in$

$10\,in > l > 2\,in$

Possible choices: 5 inches.

(vii) $x = 10\,in, y = 4\,in$

$14\,in > l > 6\,in$

Possible choices: None.

(viii) $x = 6\,in, y = 5\,in$

$11\,in > l > 1\,in$

Possible choices: 2 inches, 4 inches, 10 inches.

(ix) $x = 10\,in, y = 5\,in$

$15\,in > l > 5\,in$

Possible choices: 6 inches.

(x) $x = 10\,in, y = 6\,in$

$16\,in > l > 4\,in$

Possible choices: 5 inches.

Possible options of this exercise: 1) (2, 4, 5), 2) (4, 5, 6), 3) (5, 6, 10), 4) (1, 2, 11), 5) (1, 4, 11), 6) (1, 10, 11)

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3 years ago

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Emmett gathered some data to compare the monthly cost of renting a one-bedroom apartment in Dallas and Austin. The data collecte

Goryan [66]

Answer:

The correct options are;

a) The city's data has the higher median monthly apartment cost of the two cities by $155

b) The city's data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data

Step-by-step explanation:

The given data for the cost of renting a one-bedroom apartment in Dallas is presented as follows;

$994, $1,322, $1,075, $1,189, $1,172, $1,465, $1,215, $930, $1,090, $1,288

Which can be arranged in increasing order and analyzed to find the interquartile range, mean and standard deviation using Microsoft Excel as follows;

$930, $994, $1,075, $1,090, $1,172, $1,189, $1,215, $1,288, $1,322, and $ 1,465

The first quartile, Q₁ = $1,054.75

The third quartile, Q₃ = $1,296.5

The interquartile range = Q₃ - Q₁ = $241.75

The median = $1,180.5

The average monthly cost = $1,174

The standard deviation of the sample = $159.9597

The given data for the cost of renting a one-bedroom apartment in Austin is presented in increasing order using Microsoft Excel as follows;

$900, $950, $1,100, $1,250, $1,296, $1,375, $1,389, $1,400, $1,450, $1,495

The interquartile range, mean and standard deviation are found using Microsoft Excel as follows;

The first quartile, Q₁ = $1,062.5

The third quartile, Q₃ = $1,412.5

The interquartile range = Q₃ - Q₁ = $1,412.5 - $1,062.5 = $350

The median = $1,335.5

The average monthly cost = $1,260.5

The standard deviation of the sample = $209.0945

The difference in the median cost of the two cities is $1,335.5 - $1,180.5 = $155

Therefore, the Austin's city data has the higher median monthly apartment cost of the two cities by $155

b) The difference in the sample standard deviation of the two cities is $209.0945 - $159.9597 = $49.13476

Therefore, the Austin city data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data.

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3 years ago

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Dafna1 [17]

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3 years ago

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zepelin [54]

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