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Annette [7]
3 years ago
9

Maggie is displaying paperback and hardback books on her book shelf. The ratio of hardback books to total books is 5 to 13. Whic

h number could be the number of paperback books Maggie put on her shelf?
Mathematics
1 answer:
lubasha [3.4K]3 years ago
8 0

Answer:

Total Paperback books = 8x , where x is any positive number

Step-by-step explanation:

Let

The total books are = 13x   ; where x is any positive number

As given,

The ratio of hardback books to total books is 5 to 13.

⇒Hardback books = \frac{5}{13}× Total books

⇒Hardback books = \frac{5}{13}× 13x = 5x

So, Paperback books = Total books - Hardback books

                                    = 13x - 5x

                                    = 8x

⇒Total Paperback books = 8x

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7 0
3 years ago
f the price charged for a candy bar is​ p(x) cents, where p (x )equals 162 minus StartFraction x Over 10 EndFraction ​, then x t
andreev551 [17]

Answer:

a. 1620-x^2

b. x=810

c. Maximum value revenue=$656,100

Step-by-step explanation:

(a) Total revenue from sale of x thousand candy bars

P(x)=162 - x/10

Price of a candy bar=p(x)/100 in dollars

1000 candy bars will be sold for

=1000×p(x)/100

=10*p(x)

x thousand candy bars will be

Revenue=price × quantity

=10p(x)*x

=10(162-x/10) * x

=10( 1620-x/10) * x

=1620-x * x

=1620x-x^2

R(x)=1620x-x^2

(b) Value of x that leads to maximum revenue

R(x)=1620x-x^2

R'(x)=1620-2x

If R'(x)=0

Then,

1620-2x=0

1620=2x

Divide both sides by 2

810=x

x=810

(C) find the maximum revenue

R(x)=1620x-x^2

R(810)=1620x-x^2

=1620(810)-810^2

=1,312,200-656,100

=$656,100

7 0
4 years ago
Estimate the student's walking pace, in steps per minute, at 3:20 p.m. by averaging the slopes of two secant lines from part (a)
ehidna [41]

This question is incomplete, the complete question is;

A student bought a smart-watch that tracks the number of steps she walks throughout the day. The table shows the number of steps recorded (t) minutes after 3:00 pm on the first day she wore the watch.

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

a) Find the slopes of the secant lines corresponding to the given intervals of t.

1) [ 0, 40 ]

11) [ 10, 20 ]

111) [ 20, 30 ]

b) Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part (a). (Round your answer to the nearest integer.)

Answer:

a)

1) for [ 0, 40 ], slope is 96

11) for [ 10, 20 ],  slope is 86.3

111) for  [ 20, 30 ], slope is 116.4

b) the student's walking pace is 101 per min

Step-by-step explanation:

Given the data in the question;

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

SLOPE OF SECANT LINES

1) [ 0, 40 ]

slope =  ( 7,128 - 3,288 ) / ( 40 - 0

= 3840 / 40 = 96

Hence slope is 96

11)  [ 10, 20 ]

slope = ( 5,522 - 4,659 ) / ( 20 - 10 )

= 863 / 10 = 86.3

Hence slope is 86.3

111)  [ 20, 30 ]

slope = ( 6,686 - 5,522 ) / ( 30 - 20 )

= 1164 / 10 = 116.4

Hence slope is 116.4

b)

Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part .

Since this is recorded after 3:00 pm

{ 3:20 - 3:00 = 20 }

so t = 20 min

so by average;

we have ( [ 10, 20 ] + [ 20, 30 ] ) /2

⇒ ( 86.3 + 116.4 ) / 2

= 202.7 /2

= 101.35 ≈ 101

Therefore, the student's walking pace is 101 per minutes

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Answer:

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