Mathematics, 07.07.2020 21:01 deaishaajennings123

A camera shop stocks eight different types of batteries, one of which is type A76. Assume there are at least 30 batteries of each type. Required:
a. How many ways can a total inventory of 30 batteries be distributed among the eight different types.
b. How many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory must include at least four A76 batteries?
c. How many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory includes at most three A7b batteries?

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Answer from: bultron1966

a. 10295472 ways

b. 4272048 ways

c. 6023424 ways

Step-by-step explanation:

Given that:

Camera shop stocks ----- 8 different types of batteries

one of which is ---- A76

Assume that there are ------ at least 30 batteries of each type.

How many ways can a total inventory of 30 batteries be distributed among the eight different types.

The number of ways a total inventory of 30 batteries be distributed is :

$= \left \{ {{30+8-1} \atop {30}} \right. \}$

$= \left \{ {{37} \atop {30}} \right. \}$

$=\dfrac{37!}{30! *7!}$

$= \dfrac{37*36*35*34*33*32*31*30!}{30!*7*6*5*4*3*2*1}$

$= \dfrac{37*36*35*34*33*32*31}{7*6*5*4*3*2*1}$

= 10295472 ways

How many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory must include at least four A76 batteries?

If we must include 4 A76 batteries; then the number of ways a total inventory of 30 batteries can be distributed among eight different types of batteries will be:

30 - 4 = 26 batteries

Now;

$= \left \{ {{26+8-1} \atop {26}} \right. \}$

$= \left \{ {{33} \atop {26}} \right. \}$

$=\dfrac{33!}{26! \ \ 7!}$

$=\dfrac{33*32*31*30*29*28*27*26!}{26! \ * \ 7*6*5*4*3*2*1}$

$=\dfrac{33*32*31*30*29*28*27}{ \ 7*6*5*4*3*2*1}$

= 4272048 ways

c. If we must include at most three A7b batteries. the number of ways that a total inventory of 30 batteries can be distributed among eight different types of inventory is:

$= \sum \limits ^3 _{x=0} \left \{ {{(30-x)+7-1} \atop {30-x}} \right. \} \\ \\ \\ = \sum \limits ^3 _{x=0} \left \{ {{(30-0)+7-1} \atop {30-0}} \right. \} = (^{36}_{30})+(^{35}_{29})+ (^{34}_{28})+ (^{33}_{27})$

$= \dfrac{36!}{30! * 6!} + \dfrac{35!}{29! * 6!} + \dfrac{34!}{28! * 6!} + \dfrac{33!}{27! * 6!}$

= 1947792 + 1623160 + 1344904 + 1107568

= 6023424 ways

Answer from: Quest

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

$y\propto x$

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

Answer from: Quest

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

step-by-step explanation:

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