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trapecia [35]
2 years ago
14

If 1 cm represents 30 km on a map and colorado is shown by a 20cm by 15 cm rectangle on the map, what is the approximate actual

area of colorado
Mathematics
1 answer:
zubka84 [21]2 years ago
3 0

Answer:

20*30km = 600km

15*30km = 450km

600km * 450km = 270000 km^2

Step-by-step explanation:

assuming 1cm --- 30km causes that 2cm --- 60km, 3cm --- 90km, 4cm --- 120km etc. Notice that number of km's is 30 times greater than the number of cm's on a map

using this logic, you calculate both real dimensions of Colorado by multiplying 20*30 = 600 km and 15*30 = 450km

Then we use a formula of an area of a rectangle A=a\cdot b assuming a=600km, \ b=450km and get A=600\cdot 450=270000km^2 which is an actual area of Colorado

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If a lineman can install 12 insulators in 183/4 hours, how many insulators should he be able to install in 281/8 hours?
katrin2010 [14]

Answer: 9

Step-by-step explanation:

12 insulators = 183/4 hours

x = 281/8

x × 183/4 = 12 ×281/ 8

183x/4 = 6 × 281/4

183x / 4 = 1686 / 4

4 cancel out 4

183x = 1686

Divide bothside by 183

x = 1686/183

x= 9.21311

x = 9 approximately.

3 0
2 years ago
Dakari makes 2 dozen jars of jam. The total weight of all of the filled jars is 211.2 ounces. When empty, each jar weighs 1.2 ou
Rzqust [24]

Answer:

Step-by-step explanation:

Total weight of all of the filled jars with jam = 211.2 ounces

When empty, each jar weighs 1.2 ounces

Total weight of 2 dozens of empty jar = 24 × 1.2 ounces

= 28.8 ounces

Total weight of jam in 2 dozens jar = Total weight of all of the filled jars with jam - Total weight of 2 dozens of empty jar

= 211.2 ounces - 28.8 ounces

= 182.4 ounces

The weight of jam in one jar = 182.4 ounces / 24 jars

= 7.6 ounces

The weight of jam in one jar = 7.6 ounces

3 0
2 years ago
Choose the correct item from each drop-down menu to factor the trinomial 6y2 + 19y + 15 by grouping.
Alex73 [517]

Answer: (3y - 5) • (2y - 3)

Step-by-step explanation:      6y2 - 10y - 9y - 15

2.1     Factoring  6y2-19y+15

The first term is,  6y2  its coefficient is  6 .

The middle term is,  -19y  its coefficient is  -19 .

The last term, "the constant", is  +15

5 0
2 years ago
f1(x) = ex, f2(x) = e−x, f3(x) = sinh(x) g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the int
eimsori [14]

Answer:

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

Step-by-step explanation:

Given

f1(x) = e^x

f2(x) = e^(-x)

f3(x) = sinh(x)

g(x) = 0

We want to solve for C1, C2 and C3, such that

C1f1(x) + C2f2(x) + C3f3(x) = g(x)

That is

C1e^x + C2e^(-x) + C3sinh(x) = 0

The hyperbolic sine of x, sinh(x), can be written in its exponential form as

sinh(x) = (1/2)(e^x + e^(-x))

So, we can rewrite

C1e^x + C2e^(-x) + C3sinh(x) = 0

as

C1e^x + C2e^(-x) + C3(1/2)(e^x + e^(-x)) = 0

So we have

(C1 + (1/2)C3)e^x + (C2 + (1/2)C3)e^(-x) = 0

We know that

e^x ≠ 0, and e^(-x) ≠ 0

So we must have

(C1 + (1/2)C3) = 0...........................(1)

and

(C2 + (1/2)C3) = 0..........................(2)

From (1)

2C1 + C3 = 0

=> C3 = -2C1.................................(3)

From (2)

2C2 + C3 = 0

=> C3 = -2C2................................(4)

Comparing (3) and (4)

2C1 = 2C2

=> C2 = C1

Let C1 = C2 = K

C3 = -2K

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

3 0
2 years ago
Printed circuit cards are placed in a functional test after being populated with semiconductor chips. A lot contains 140 cards,
hjlf

Answer:

a) 0.9644 or 96.44%

b) 0.5429 or 54.29%

Step-by-step explanation:

a) The probability that at least 1 defective card is in the sample P(A) = 1 - probability that no defective card is in the sample P(N)

P(A) = 1 - P(N) .....1

Given;

Total number of cards = 140

Number selected = 20

Total number of defective cards = 20

Total number of non defective cards = 140-20 = 120

P(N) = Number of possible selections of 20 non defective cards ÷ Number of possible selections of 20 cards from all the cards.

P(N) = 120C20/140C20 = 0.0356

From equation 1

P(A) = 1 - 0.0356

P(A) = 0.9644 or 96.44%

b) Using the same method as a) above

P(A) = 1 - P(N) .....1

Given;

Total number of cards = 140

Number selected = 20

Total number of defective cards = 5

Total number of non defective cards = 140-5 = 135

P(N) = 135C20/140C20 = 0.457

From equation 1

P(A) = 1 - 0.4571

P(A) = 0.5429 or 54.29%

8 0
2 years ago
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