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2 years ago

The area A of the rectangle shown is described with the inequality 100 ≤ A ≤ 1,000. Write and solve a compound inequality for x.

Mathematics

2 answers:

frutty [35]2 years ago

8 0

Answer:

The correct answer is 1.5 to 5.5

Step-by-step explanation:

kipiarov [429]2 years ago

5 0

Answer:

1.5 to 5.5 not viable

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HURRY PLEASE When 5 and 6 are multiplied by the same factor, how do the ratios compare to the ratio 5:6? The ratios are equivale

erma4kov [3.2K]

Answer:

The ratios are equivalent.

Step-by-step explanation:

If you multiply both 5 and 6, the ratio stays the same.

5:6 = 10:12 = 15:18 = 20:24 etc.

You can try this by dividing the larger number by the smaller number. You'll always get 1.2 as the "relationship", or ratio, is preserved.

24/20 = 1.2

15/18 = 1.2

10/12 = 1.2

6/5 = 1.2

6 0

2 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0

2 years ago

Mr. Peter charges $2.25 for the usage of laptop for 9 hours. How much Victor needs to pay for 15 hours of usage?

Papessa [141]

Answer:

3.75

Step-by-step explanation:

2.25 divided by 9

= 0.25

then

0.25 * 9

= 3.75

5 0

2 years ago

Read 2 more answers

louise bought 80 metres of fencing to make an enclosure for her pet dog .tommy if louise expects a rectangular enclosure what is

OleMash [197]

Answer:

400 m^2.

Step-by-step explanation:

The largest area is obtained where the enclosure is a square.

I think that's the right answer because a square is a special form of a rectangle.

So the square would be 20 * 20 = 400 m^2.

Proof:

Let the sides of the rectangle be x and y m long

The area A = xy.

Also the perimeter 2x + 2y = 80

x + y = 40

y = 40 - x.

So substituting for y in A = xy:-

A = x(40 - x)

A = 40x - x^2

For maximum value of A we find the derivative and equate it to 0:

derivative A' = 40 - 2x = 0

2x = 40

x = 20.

So y = 40 - x

= 40 - 20

=20

x and y are the same value so x = y.

Therefore for maximum area the rectangle is a square.

3 0

2 years ago

A rectangular piece of land is 40m long and 25m wide. A path of uniform width and 426 m^2 area sorrounds

fredd [130]

Answer:

Width of the uniform path that surrounds the piece of land = 3 m

Step-by-step explanation:

Let the width of the path that surrounds the piece of land be x.

The situation described is sketched in the attached image to this solution.

The dimension of the Length of the piece of land including the uniform path that surrounds it = (40 + 2x) m

The Breadth of the piece of land including the uniform path that surrounds it = (25 + 2x) m

The area of the piece of land including the uniform path that surrounds it

= (Area of the piece of land) + (Area of the uniform path that surrounds it)

Area of the piece of land = Length × Breadth = 40 × 25 = 1000 m²

Area of the uniform path that surrounds it = 426 m² (Given in the question)

The area of the piece of land including the uniform path that surrounds it = 1000 + 426 = 1426 m²

But the area of the piece of land including the uniform path that surrounds it is also

= (Length of the piece of land including the uniform path that surrounds it) × (Breadth of the piece of land including the uniform path that surrounds it

= (40 + 2x) + (25 + 2x)

= 1000 + 80x + 50x + 4x²

= (4x² + 130x + 1000) m²

Equation these 2 areas

4x² + 130x + 1000 = 1426

4x² + 130x - 426 = 0

Solving the quadratic equation

x = 3 or -35.5

Since the width cannot be negative,

x = 3 m

Hope this Helps!!!

7 0

2 years ago