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3 years ago

A square with sides of length 5 is positioned inside a square with sides of length 7. What is

Mathematics

1 answer:

kow [346]3 years ago

3 0

Answer: 24cm²

Step-by-step explanation:

Let square with length 5cm be represented with A, and square with 7cm be represented with B.

Area of a square is L²

Square A will have an area of 5² = 25cm²

Square B will have an area of 7² = 49cm²

Total area between the two squares will be = Area of Square (B - A)

i.e.

Total Area = 49cm² - 25cm² = 24cm²

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A triangular pyramid has a base shaped like an equilateral triangle. The legs of the equilateral triangle are all 4 feet long, a

9966 [12]

Answer:

Total surface area of the pyramid = $25ft^{2}$

Step-by-step explanation:

To find the surface area of the triangular- based pyramid, we will use the formula:

Surface area = Area of base + Area of the three faces

Area of base of the pyramid = Area of equilateral triangle = $0.5 \times base \times height$

$= 0.5 \times 4 \times 3.5= 7ft ^{2}$

Area of the slant faces of the pyramid is $0.5 \times base \times slant height$

$0.5 \times 4 \times 3 = 6 ft^{2}$

Since there are 3 slant surfaces the total surface area of the pyramid = $7 + 3 \times (6)= 25ft^{2}$

6 0

4 years ago

What is the sales of volume V, of a particular product with a net income of $5000 given a sales price of $40, a variable cost of

iVinArrow [24]

Answer:

The sales of volume V, of a particular product with a net income of $5000 given a sales price of $40, a variable cost of $15?and $1000 in fixed cost is 240

Step-by-step explanation:

Given:

Net income= $5000

sales price= $40

variable cost =$15 and $1000

To Find:

sales of volume V=?

Solution:

We Know that

$\text {Income}=40 * x$

Where x= number of units sold

Similarily,

$\text { Net Income }=40 \mathrm{x}-\text { VARIABLE } \cos \mathrm{T} * \mathrm{x}-1000$

Substituting the known values,

$\text { Net Income }=40 \mathrm{x}-\text{VARIABLE COST} * \mathrm{x}-1000$

$5000=40 \mathrm{x}-15 * \mathrm{x}-1000$

Sloving the equation,

$5000=40 \mathrm{x}-15 \mathrm{x}-1000$

$5000=25 \mathrm{x}-1000$

$5000+1000=25 \mathrm{x}$

$6000=25 \mathrm{x}$

$x=\frac{6000}{25}$

$x=240$

5 0

3 years ago

Combine like terms to create an equivalent expression. 3.26d+9.75d-2.65

earnstyle [38]

Add 3.26d+9.75-2.65

13.01y - 2.65

6 0

3 years ago

Read 2 more answers

1. Which economic policy was used to help American industry during the Second Industrial Revolution?

enot [183]

Hello!

Economic policy used in the US during the second industrial revolution was the model of laissez-faire capitalism. Socialism and populism have not been used often in the US.

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The main consequence of the use of holding was to reduce the power of couplings, since it is configured symbolic monopoly.

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The steel industry has contributed to the territorial and technological expansion. This is remarkable when we look at the railroads.

hugs!

5 0

3 years ago

Read 2 more answers

The probability that a single radar station will detect an enemy plane is 0.65.

taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

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 combinatios 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.

In this problem, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that $n = 0.65$ .

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which $P(X = 0) \leq 0.02$ .

n = 1.

$P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35$

n = 2

$P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225$

n = 3

$P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429$

n = 4

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

$E(x) = np$

So when $n = 7$

$E(x) = 7*(0.65) = 4.55$

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

6 0

3 years ago