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2 years ago

11

The rectangular floor of a classroom is 32 feet in length and 36 feet in width. A scale drawing of the floor has a length of 8 i

nches. What is the perimeter, in inches, of the floor in the scale drawing?

1 answer:

Vanyuwa [196]2 years ago

6 0

Answer:

34

Step-by-step explanation:

32 / 8 = 4
36 / 4 = 9

perimeter of rectangle = 2(Length) + 2(width)
P = 2(8) + 2(9)
P = 16 + 18
P = 34

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The Alliance Corp. expects to sell the following number of units of copper cables at the prices indicated, under three different

Serggg [28]

Answer:

$E(x) = 21834$

Step-by-step explanation:

Given

The above data

Required

Determine the expected value E(x)

E(x) is calculated as thus;

$E(x) = \sum xf(x)$

Where f(x) represents the total sale of each outcome;

For A:

$f(x) = 295 *34$

$f(x) = 10030$

For B:

$f(x) =510 *49$

$f(x) =24990$

For C:

$f(x) = 790 *59$

$f(x) = 46610$

Hence, E(x):

$E(x) = 0.5 * 10030 + 0.30 * 24990 + 0.20 * 46610$

$E(x) = 21834$

8 0

3 years ago

The fish population is decreasing at a rate of 3% per year. In 2002 there were about 2100 fish. Find the fish population in 2010

djyliett [7]

Answer:

Population of fish in 2010 = 1,646 fish (Approx.)

Step-by-step explanation:

Given:

Population of fish in 2002 = 2100 fish

Decrease rate = 3% per year

Find:

Population of fish in 2010

Computation:

Number of year = 2010 - 2002

Number of year = 8 year

A = P[1+r]ⁿ

Population of fish in 2010 = Population of fish in 2002[1+(-3%)]ⁿ

Population of fish in 2010 = 2100[1-0.03]⁸

Population of fish in 2010 = 2100[0.97]⁸

Population of fish in 2010 = 2100[0.7837]

Population of fish in 2010 = 1645.77

Population of fish in 2010 = 1,646 fish (Approx.)

4 0

3 years ago

The quantity demanded x each month of Russo Espresso Makers is 250 when the unit price p is $138. The quantity demanded each mon

Stells [14]

Answer:

(a) $D(q)=\frac{-1}{25} q+148$

(b) $S(q)=\frac{1}{50}q+58$

(c) $p_{*} =88\\\\q_{*} =1500$

Step-by-step explanation:

(a) For the demand equation D(q) we have

<em>P1: 138 Q1: 250</em>

<em>P2: 108 Q2: 1000</em>

We can find <u><em>m</em></u>, which is the slope of the demand equation,

$m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{108-38}{1000-250} =\frac{-30}{750}=\frac{-1}{25}$

and then we find b, which is the point where the curve intersects the y axis.

We can do it by plugging one point and the slope into the line equation form:

$y=mx+b\\\\D(q)=mq+b\\\\138=\frac{-1}{25}(250) +b\\\\138=-10+b\\\\138+10=b=148$

<em>With b: 148 and m: -1/25 we can write our demand equation D(q)</em>

$D(q)=\frac{-1}{25} q+148$

(b) to find the supply equation S(q) we have

<em>P1: 102 Q1: 2200</em>

<em>P2: 102 Q2: 700</em>

<em></em>

Similarly we find <em>m</em>, and <em>b</em>

$m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{72-102}{700-2200} =\frac{-30}{-1500}=\frac{1}{50}$

$y=mx+b\\\\D(q)=mq+b\\\\72=\frac{1}{50}(700) +b\\\\72=14+b\\\\72-14=b=58\\$

<em>And we can write our Supply equation S(q):</em>

$S(q)=\frac{1}{50}q+58$

(c) Now we may find the equilibrium quantity q* and the equilibrium price p* by writing <em>D(q)=S(q)</em>, which means the demand <u><em>equals</em></u> the supply in equilibrium:

$D(q)=S(q)\\\\\frac{-1}{25}q+148=\frac{1}{50}q+58\\\\$

$148-58=\frac{1q}{50} +\frac{1q}{25} \\\\90= \frac{1q}{50} +\frac{2q}{50}\\\\90=\frac{3q}{50}\\ \\q=1500\\\\$

We plug 1500 as q into any equation, in this case S(q) and we get:

$S(q)=\frac{1}{50}q+58\\\\S(1500)=\frac{1}{50}(1500)+58\\\\S(1500)=30+58\\\\S(1500)=88$

Which is the equilibrium price p*.

8 0

3 years ago

Find the solution <br> Y = -3x - 5

igomit [66]

Answer:
-5/3

Explanation:

5 0

2 years ago

Pls help me frens......

Natalija [7]

Answer:

V= 8

x=4.5

BCF =48 CFE=132

V=352.8 =8.4x8.4x5

For the volume one i cant read the number but if the bottom number is 8 then V=803.84

if its 6 then v = 452.16

Step-by-step explanation:

v= 2x2x2 v = 8

x is 11x-9=7x+9 | 4x=18 | x=4.5

15x=180 x=12 | CFE = 11x12=132 | BCF = 4x12=48

4 0

3 years ago