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2 years ago

Find the unit rate. 5/8 mile in 1/4 hour

Mathematics

2 answers:

ivolga24 [154]2 years ago

8 0

Answer:

5/8 : 1/4 :: m : 1 proportion

5/8 = m/4 product means/extremes

5/2 = m

Step-by-step explanation:

11111nata11111 [884]2 years ago

4 0

2.5 mph is the correct answer take it from an ph expert

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What is the area of a circle with the diameter of 8

Mars2501 [29]

Area = πr^2
π4^2 =16π
=50.3

5 0

3 years ago

Read 2 more answers

Unsure how to do this calculus, the book isn't explaining it well. Thanks

krok68 [10]

One way to capture the domain of integration is with the set

$D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}$

Then we can write the double integral as the iterated integral

$\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx$

Compute the integral with respect to $y$ .

$\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)$

Compute the remaining integral.

$\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}$

We could also swap the order of integration variables by writing

$D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}$

and

$\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy$

and this would have led to the same result.

$\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)$

$\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)$

7 0

1 year ago

11+3(2x-1) = 20 -- x

attashe74 [19]

Answer:

x=3.50

Step-by-step explanation:

2*3.50=7

7-1=6

3+6=9

9+11=20

4 0

2 years ago

What is the concentration of H+ ions at a pH = 2?

cestrela7 [59]

Answer:

0.01

0.000001

10,000

Step-by-step explanation:

i got it on edge

5 0

1 year ago

The following data represent the pulse rates? (beats per? minute) of nine students enrolled in a statistics course. Treat the ni

charle [14.2K]

Answer:

(a)77.4bpm

(b)Mean of Sample 1 = 70.3 beats per minute.

Mean pulse of sample 2 = 70 beats per minute.

(c)

The mean pulse rate of sample 1 underestimates the population mean.
The mean pulse rate of sample 2 underestimates the population mean.

Step-by-step explanation:

(a)Population mean pulse.

The pulse of the nine students which represent the population are:

Perpectual Bempah 64
Megan Brooks 77
Jeff Honeycutt 89
Clarice Jefferson 69
Crystal Kurtenbach 89
Janette Lantka 65
Kevin McCarthy 88
Tammy Ohm 69
Kathy Wojdya 87

$\text{Population Mean} =\dfrac{64+77+89+69+89+65+88+69+87}{9} \\=\dfrac{697}{9} \\\\=77.44$

The population mean pulse is approximately 77.4 beats per minute.

(b)Sample 1: {Janette,Clarice,Megan}

Janette: 65bpm
Clarice: 69bpm
Megan: 77bpm

Mean of Sample 1

$\text{Sample 1 Mean} =\dfrac{65+69+77}{3} \\=\dfrac{211}{3} \\\\=70.3$

Sample 2: {Janette,Clarice,Megan}

Perpetual: 64bpm
Clarice: 69bpm
Megan: 77bpm

Mean of Sample 2

$\text{Sample 2 Mean} =\dfrac{64+69+77}{3} \\=\dfrac{210}{3} \\\\=70$

The mean pulse of sample 1 is approximately 70.3 beats per minute.

The mean pulse of sample 2 is approximately 70 beats per minute.

(c)

The mean pulse rate of sample 1 underestimates the population mean.
The mean pulse rate of sample 2 underestimates the population mean.

7 0

2 years ago