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Anastaziya [24]

3 years ago

8

Kristin owns a bakery called Kristin’s cakes and n’ such and is considering lowering the price of her cakes. Kristen polls her c

ustomers and determines that she can sell 100 cakes each week when she changes $25 each. She also discovers that every $1 decrease in price of the cake, she will sell 5 more cakes. The portion of the graph below represents kristen’s projected weekly revenue for each decrease in price.
Which of the following are true? Select all that apply.

2 answers:

den301095 [7]3 years ago

8 0

1,3,5
4 looks like true but decreasing but 2 dollars also gives max revenue, so I not sure about it

AleksAgata [21]3 years ago

3 0

Answer:

If Kristin does not decrease the price of her cakes,her projected weekly revenue from cake sales will be $ 2500.

Since if she is not decreasing her price that means the point will be on the Revenue axis.

and the point is 2500.

Hence, the statement is True.

If Kristin decreases the price of her cakes, her projected weekly revenue will also decrease.

This statement is False as while decreasing the price from $ 2 to $ 3 the revenue remains same.

If Kristin decreases the price of her cake by $ 1, her projected weekly revenue will be $ 2520.

This statement is True.

Since when there is a decrease in $ 1 this means that 5 cakes increases and price per cake is $ 24.

i.e. 105 cakes are being sold at a rate of $ 24 each.

Hence, the total revenue is:

24×105=$ 2520.

Kristin should decrease the price of her cake by $ 3 ,in order to obtain the maximum weekly revenue.

This statement is False.

Since, the maximum revenue is obtained when there is no decrease.

Kristin will receive the same revenue if she sells the cake for $ 24 or $ 21.

This statement is True.

Since when she sells for $ 24 each she has to sell 105 cakes.

so, revenue obtained is:

24×105=$ 2520.

When she sells for $ 21, she has to sell 120 cakes and the revenue obtained is:

20×120=$ 2520.

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Carnival T charges an entrance fee of $7.00 and $0.50 per ticket for the rides. Carnival Q charges an entree fee of $12.00 and $

FrozenT [24]

Answer:

$30

Step-by-step explanation:

Calculation for the tickets that must be purchased for Carnival T and Carnival Q to be the same

Based on the information given let x be the number of ticket to be purchased .

Carnival T entrance fee= $7.00

Ride= $0.50 per ticket

Carnival Q entree fee =12.00

Ride= $0.25 per ticket

Tickets=$7.00 + ($0.50* x) = $12.00 + ($0.25* x)

.25 x = 5.00

Hence:

x=$.25+$5.00

×=$30

Therefore the amount of tickets that must be purchased in order for the total cost at Carnival T and Carnival Q to be the same will be $30

3 0

3 years ago

Me podrían ayudar y explicarme el proceso please

kakasveta [241]

Answer: Encontrarás las respuestas debajo de cada explicación. Espero que consideres darme brainliest y 5 estrellas, y más importante, que hayas entendido.

Step-by-step explanation:

Esto es la regla de 3. Conoces 3 valores y necesitas encontrar un 4to valor.

Ya sabemos que 1 caja pesa 20kg,

1 - 20kg

ahora queremos saber cuantas cajas (x) equivalen a 7653kg

x - 7653kg

Hacemos un pequeño cuadro poniendo en un lado los numeros de cajas y del otro el peso.

Pongamos la cantidad de cajas en la izquierda y el peso en la derecha

$\frac{1}{x} =\frac{20kg}{7653kg}$

Procedemos a multiplicar en cruz. Solo podemos multiplicar si tenemos ambos valores. En este caso, los valores en cruz que tenemos son 1 y 7653kg porque en la otra cruz (20 y x) tenemos nuestra incognita. Y, por ultimo, dividimos entre el numero cuya cruz es con la incognita (20kg)

Esto nos deja la expresión así;

$x=\frac{1*7653kg}{20kg}$

$x=382.65$

Como una caja no puede ser un numero decimal, redondeamos.

$x=383$

Por lo tanto, se necesitan aproximadamente 383 cajas para que su peso equivalga a 7653 kg.

-------------------------------------------------------------------------------------------------------

La otra pregunta dice: Si una caja pesa 20kg, (1 - 20kg), ¿Cuántas cajas se necesitan para equivaler 9500kg? (x - 9500kg)

Hacemos lo mismo que arriba, ponemos la cantidad de cajas de un lado, y el peso de otro lado.

$\frac{1}{x}=\frac{20kg}{9500kg}$

Multiplicamos aquellos valores en cruz que conozcamos y dividimos por el valor cuya cruz sea con x.

$x=\frac{1*9500kg}{20kg}$

$x=475$

Esta es la cantidad de cajas que se necesitan para que su pesa equivalga a 9500kg.

--------------------------------------------------------------------------------------------------------

Por último, la pregunta dice, ¿Cuántos kilogramos hay en 873 cajas?

Ya sabemos que una caja pesa 20kg ( 1 - 20kg ) y ahora buscamos cuantos kg hay en 873 cajas; es decir, (873 - x)

Nos quedaría así;

$\frac{1}{873}=\frac{20kg}{x}$

Ahora, nuestra cruz es 873 y 20kg, y el divisor es 1.

$x=\frac{873*20kg}{1}$

$x=17460kg$

Este sería el peso de 873 cajas.

5 0

3 years ago

WORTH 100 POINTS!

zzz [600]

Answer:

7miles

equations shown in work image below

work shown on the paper with more explanations because text is hard to write equations (sorry I used km but it is miles for your problem.)

:) please give a heart for thanks

and brainliest crown :)

Step-by-step explanation:

For both d=r*t

Each bikes to others house and leave at the same time so for Tony,

11 = 30*t

t= 11/30 hours

for Gerry,

11 = 25*t

t= 11/25 hours

Comparing these, with LCD,

Tony's time is 55/150

Gerry's time is 66/150

So Tony got to Gerry's house 11/150 hours quicker than Gerry got to Tony's house.

Now they turn around and head back toward each other. If they leave at the same exact time, the elapsed time for each until they met would be the same so t (for tony) = t (for gerry), but Tony got a head start.

We start the stopwatch when Gerry turns around 11/150 hours after Tony has turned around. If Gerry's time to where they meet from Tony's house t(gerry), then Tony's time from Gerry's house is t(gerry) + 11/150

5 0

3 years ago

Read 2 more answers

What number has exactly 6 factors

RoseWind [281]

The factors of 12 are 1, 2, 3, 4, 6, and 12 .

The factors of 32 are 1, 2, 4, 8, 16, and 32 .

4 0

3 years ago

Read 2 more answers

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$

And the magnitude of the derivative of unit tangent vector is

$||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2$

The curvature is

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1$

The tangential component of acceleration is given by the formula

$a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}$

We know that $\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ and $||\boldsymbol{r}'(t)}||=2$

$\frac{d}{dt}\left(2\right)\: = 0$ so

$a_{\boldsymbol{T}}=0$

The normal component of acceleration is given by the formula

$a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2$

We know that $\kappa=1$ and $\frac{ds}{dt}=2$ so

$a_{\boldsymbol{N}}=1 (2)^2=4$

3 0

3 years ago