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2 years ago

What is the area please help me

Mathematics

1 answer:

Angelina_Jolie [31]2 years ago

5 0

The area and perimeter of the picture is 576 inches square and 96 inches respectively

<h3>How to determine the area</h3>

Given that each frame is a rectangle and four frames makes up the picture

Note that the picture takes the shape of a square

Let's find the total perimeter

Perimeter of the picture = sum of the four rectangular frame perimeters

Perimeter = 24 + 24 + 24 + 24

Perimeter = 96 inches

Formula for area of a square = a^2

Where 'a' is the length of the side = 24 inches

Substitute the value of 'a'

Area of the picture = 24^2

Area of the picture = 576 inches square

Thus, the area and perimeter of the picture is 576 inches square and 96 inches respectively.

Learn more about a square here:

brainly.com/question/24487155

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Write six and five hundredths in decimal form.

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6.05 is six and five hundredths in decimal form.

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3 years ago

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Wats 3/500 as a percent

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Your answer is 0.6%.

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3 years ago

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A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade

Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

$P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}$

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

$P(x\geq1)=1-P(x$

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly n= 5

Number of blades in an assembly dull is M = 10.

The probability mass function is,

$P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}$

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

$P(x\geq 1)=1- P(x$

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly N = 5

Number of blades in an assembly dull is M = 10

From the information,

The probability that assembly is replaced (P) is 0.7069.

The probability that assembly is not replaced is (Q) is,

$q=1-p\\= 1-0.7069= 0.2931$

The geometric probability mass function is,

$P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)$

The probability that assembly is replaced no replaced until the third day of evaluation is,

$P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607$

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 2.

$P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005$

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 6.

$P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968$

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

$P(x\geq 1)=1- P(x$

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0

4 years ago

Assume that all triangles have interior angles less than 90°.A surveyor sights on a survey marker that is 132.3m distant. She ne

Vadim26 [7]

Answer:

85.9 m

Step-by-step explanation:

The law of sines can help figure this.

The remaining angle in the triangle is ...

180° -75° -68° = 37°

This is the angle opposite the leg from the surveyor to the second marker. Referencing the attachment, we have ...

b/sin(B) = c/sin(C)

b = sin(B)·c/sin(C) = 132.3·sin(37°)/sin(68°) ≈ 85.873 . . . meters

The surveyor is about 85.9 meters from the second marker.

4 0

3 years ago

Select the correct answer from each drop-down menu. Complete the statement. The solutions of sin2x= √3/2

alex41 [277]

Answer:

henc X = 30°

Step-by-step explanation:

here is the proof

when X=30° then

sin2x = sin2×30

=sin 60°

= √3/2

or else,

putting value of X = 30° then

sin2x= 2sinxcosx

= 2×sin30°×cos30°

=2×1/2×√3/2

= 2√3/4

= √3/2

hence proved sin2x= √3/2.

7 0

3 years ago