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3 years ago

Find the class boundaries.

Mathematics

1 answer:

miv72 [106K]3 years ago

7 0

Answer:

Step-by-step explanation:

To find the class boundary of this

We have to get the lower boundary and upper boundary

54.5-59.5

59.5-64.5

64.5-69.5

69.5-74.5

74.5-79.5

79.5-84.5

84.5-89.5

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10 pts

RoseWind [281]

263.25 is the volume of the box in inches cubes you have to multiply 4.5 by 9 by 6.5

6 0

3 years ago

Read 2 more answers

I NEED THIS ASAP, PLEASE HELP!!!

WINSTONCH [101]

Answer: 22.5 ; 5 ; 14

Step-by-step explanation:

Given the dataset:

{20,22,23,24,26,26,28,29,30}

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 9

Q1 = 1/4 (9 + 1)th term

Q1 = 1/4(10) = 2.5

We average the 2nd and 3rd term:

(22 + 23) / 2

45 / 2 = 22.5

B) The interquartile range(IQR) of the dataset :

{62,63,64,65,67,68,68,68,69,74}

IQR = Q3 - Q1

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 10

Q1 = 1/4 (10 + 1)th term

Q1 = 1/4(11) = 2.75 term

We take the average of the 2nd and 3rd term:

(63 + 64) / 2

45 / 2 = 63.5

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 10

Q3 = 3/4 (10 + 1)th term

Q3 = 3/4(11) = 8.25 term

We take the average of the 8th and 9th term:

(68 + 69) / 2

137 / 2 = 68.5

IQR = Q3 - Q1

IQR = 68.5 - 63.5

IQR = 5

C) give the dataset :

{7,8,8,9,10,12,13,15,16}

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 9

Q3 = 3/4 (9 + 1)th term

Q3 = 3/4(10) = 7.5 term

We take the average of the 7th and 8th term:

(13 + 15) / 2

28 / 2 = 14

7 0

3 years ago

The total stopping distance T for a vehicle is T = 2.5x + 0.5x 2 , where T is in feet and x is the speed in miles per hour. Appr

Kobotan [32]

Answer:

% change in stopping distance = 7.34 %

Step-by-step explanation:

The stooping distance is given by

$T = 2.5 x + 0.5 x^{2}$

We will approximate this distance using the relation

$f (x + dx) = f (x)+ f' (x)dx$

dx = 26 - 25 = 1

T' = 2.5 + x

Therefore

$f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x$

This is the stopping distance at x = 25

Put x = 25 in above equation

2.5 × (25) + 0.5× $25^{2}$ + 2.5 + 25 = 402.5 ft

Stopping distance at x = 25

T(25) = 2.5 × (25) + 0.5 × $25^{2}$

T(25) = 375 ft

Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft

% change in stopping distance = $\frac{27.5}{375}$ × 100

% change in stopping distance = 7.34 %

6 0

3 years ago

Look at the 2 pictures to solve

bija089 [108]

Answer:

x = 15.3

x = 20.9

Step-by-step explanation:

sin(46) = 11/x

⇒ x = 11/sin(46) = 15.3

cos(55) = 12/x

⇒ x = 12/cos(55) = 20.9

3 0

2 years ago

The formula for velocity of an object is the equals d over t where he is a velocity of the object t is the distance traveled and

butalik [34]

Answer:

$(a)\ t =\frac{d}{v}$

$(b)\ d = vt$

Step-by-step explanation:

The question is mixed up with details of another question. See comment for original question

Given

$v = \frac{d}{t}$

$v \to velocity$

$d \to distance$

$t \to time$

Solving (a): Solve for time

We have:

$v = \frac{d}{t}$

Cross multiply

$t * v = d$

Make t the subject

$t =\frac{d}{v}$

Solving (b): Solve for distance

We have:

$v = \frac{d}{t}$

Cross multiply

$d = v * t$

$d = vt$

7 0

3 years ago