If CE=7 in. and AC=19 in., AE=

Una empresa exportadora de frutas vende dos variedades de frambuesas: estándar y de lujo. Una caja de freses estándar se vende e

aksik [14]

Answer:

83.33 cajas estándar de frambuesas y 51.67 cajas de lujo de frambuesas.

Step-by-step explanation:

Representemos el número de cajas como

A = caja estándar de frambuesas

B = caja de lujo de frambuesas

Caja estándar de frambuesas = $ 7 Caja de lujo de frambuesas = 10.

A + B = 135 ......... Ecuación 1

B = 135 - A

7A + 10B = 1100 ........... Ecuación 2

Sustituir

135 - A para B en la ecuación 2

7A + 10 (135 - A) = 1100

7A + 1350 -10A = 1100

7A - 10A = 1100-1350

-3A = - 250

A = 250/3

A = 83.33 cajas

Sustituye 83.33 por A en la ecuación 1

A + B = 135

83,33 + B = 135

B = 135 - 83.33 = 51.67 cajas

Por lo tanto, vendió,

83.33 cajas estándar de frambuesas y 51.67 cajas de lujo de frambuesas.

6 0

3 years ago

mckenna and lara work for their uncle eddie, who repairs skateboards and bicycles. uncle eddie is leaving for a vacation to the

juin [17]

The <em><u>correct answer</u></em> is:

15 bicycles and 6 skateboards.

Explanation:

Let x represent the number of bicycles and y represent the number of skateboards. The first equation in our system is then

x+y = 21, since there are 21 total items in the shop.

Each bicycle has 2 tires and each skateboard has 4; this gives us the equation

2x+4y=54 (because they are ordering 54 tires).

This makes our system

$\left \{ {{x+y=21} \atop {2x+4y=54}} \right.$

We will use elimination to solve this; we first need one of the variables to have the same coefficient in both equations. We will make x the same, and we will do this by multiplying the first equation by 2:

$\left \{ {{2(x+y=21)} \atop {2x+4y=54}} \right. \\
\\\left \{ {{2x+2y=42} \atop {2x+4y=54}} \right.$

Now we will subtract the second equation from the first one:

$\left \{ {{2x+2y=42} \atop {-(2x+4y=54)}} \right. \\
\\-2y=-12$

Divide both sides by -2:

-2y/-2 = -12/-2

y = 6

There were 6 skateboards. Substituting this into the first equation,

x+6 = 21

Subtract 6 from each side:

x+6-6 = 21-6

x = 15

There were 15 bicycles.

3 0

3 years ago

The exact value of cos 13pi/8 is

Naily [24]

Answer:

answer is : Cos(13pi/8) = 0.3826

Step-by-step explanation:

We have, Cos (13pi/8)

Since 13pi/8 can be shown as 3pi/2 < 13pi/8 < 2pi

Hence 13pi/8 lies on fourth quadrant.

In fourth quadrant cosine will be positive.

Cos (13pi/8) = cos(3pi/2 + pi/8)

applying formula cos(A+B) = cos A cosB - sinAsinB

i.e Cos(3pi/2 + pi/8) = cos(3pi/2)cos(pi/8) - sin(3pi/2)sin(pi/8)

∵ Remember cos(3pi/2) =0 , sin(3pi/2) = -1

Cos(3pi/2 + pi/8) = 0 cos(pi/8) - (-1)sin(pi/8)

Cos(3pi/2 + pi/8) = 0 + 0.3826

Cos(3pi/2 + pi/8) = 0.3826

Hence we got Cos(13pi/8) = 0.3826

6 0

3 years ago

Read 2 more answers

Solve this please…..

Paraphin [41]

Answer:

17,12,15,14,12 answer 90

6 0

2 years ago

1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,

crimeas [40]

Answer:

Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

From table it is clear that the total outcomes possible are 27.

We know that, probability can be defined as,

$PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}$

a) All Members Assigned to the Same Station.

Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

$PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111$

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).

Therefore,

$PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666$

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).

Therefore,

$PROBABILITY\;(Case\;c) = \frac{6}{27}\\\\PROBABILITY\;(Case\;c) = 0.222$

3 0

3 years ago