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2 years ago

Kim bought 12 boxes of drinks

Mathematics

1 answer:

salantis [7]2 years ago

8 0

Answer:

The amount Kim sold each reduced price drink is $1.25 each

Step-by-step explanation:

The number of boxes of drinks Kim bought = 12 boxes

The amount Kim paid for each box = $15

The number of drinks in each box = 12 drinks

The amount Kim sild 3/4 of the drinks = $1.50 each

The amount Kim sold the remaining drinks = At a reduced rate

The amount of profit Kim made = 15%

Therefore, we have;

The total amount Kim bought the whole drinks = 12 × $15 = $180

The total number of drinks = 12 × 12 = 144 drinks

3/4 of the drinks = 3/4×144 = 108 drinks

The amount Kim sold the 108 drinks = 108 × $1.5 = $162

The amount of profit Kim made = 15%

Therefore;

(((The total amount Kim sold the whole drinks) - (The total amount Kim bought the whole drinks))/(The total amount Kim bought the whole drinks)) × 100 = Percentage profit

(((The total amount Kim sold the whole drinks) - ($180))/($180)) × 100 = 15%

The total amount Kim sold the whole drinks = 0.15×$180 + $180 = $207

The total number of the remainder of the drinks = 144 - 108 = 36 drinks

The amount Kim sold the remainder of the drinks = $207 - $162 = $45

The amount Kim sold each of the remainder of the drinks at reduced price = $45/36 = $1.25

Therefore, the amount Kim sold each reduced price drink = $1.25 each.

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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

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A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$

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2 years ago

A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave

erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, $L$ = 120 cm

Its wavelength $\lambda_1$ = 480 cm

$\lambda_2$ = 160 cm and $\lambda_3$ = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

The fundamental wavelength of a pipe which is open at both ends is 2L.
The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ $4L=4(120)=480\ cm$