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3 years ago

Challenge A bag contains pennies, nickels, dimes, and quarters. There are 50 coins in all. Of the

Mathematics

1 answer:

bagirrra123 [75]3 years ago

5 0

Answer:

$6.35

Step-by-step explanation:

Number of pennies = 10% of the 50 coins = .10(50) = 5

Number of dimes = 34% of the 50 coins = .34(50) = 17

Number of nickels = 7 + no. of pennies = 7 + 5 = 12

Number of quarters = 50 - 5 - 17 - 12 = 16

Each penny is 1 cent, each nickel is 5 cents, each dime is 10 cents, and each quarter is 25 cents.

Amount of money in the bag = 5(1) + 17(10) + 12(5) + 16(25) = 5 + 170 + 60 + 400 = 635 cents = $6.35

Read more about skewness at:

brainly.com/question/27701525

#SPJ1

8 0

2 years ago

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

3 years ago

Question 5 (1 point)

Leokris [45]

A: Add up to 180 degrees

8 0

3 years ago

Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6

allsm [11]

Answer:

$L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]$

Step-by-step explanation:

Given that:

$f(t) = 12 cos (t- \dfrac{\pi}{6})$

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

$f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}]$

$f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}]$

$f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)$

$L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]$

$L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]$

$L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}$

$L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}$

$L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}$

$L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]$

7 0

3 years ago