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melamori03 [73]
2 years ago
10

The following graphs have a scale assigned to them: The area of each grid

Mathematics
1 answer:
dezoksy [38]2 years ago
6 0

The graphs that are density curves for a continuous random variable are: Graph A, C, D and E.

<h3>How to determine the density curves?</h3>

In Geometry, the area of the density curves for a continuous random variable must always be equal to one (1). Thus, we would test this rule in each of the curves:

Area A = (1 × 5 + 1 × 3 + 1 × 2) × 0.1

Area A = 10 × 0.1

Area A = 1 sq. units (True).

For curve B, we have:

Area B = (3 × 3) × 0.1

Area B = 9 × 0.1

Area B = 0.9 sq. units (False).

For curve C, we have:

Area C = (3 × 4 - 2 × 1) × 0.1

Area C = 10 × 0.1

Area C = 1 sq. units (False).

For curve D, we have:

Area D = (1 × 4 + 1 × 3 + 1 × 2 + 1 × 1) × 0.1

Area D = 10 × 0.1

Area D = 1 sq. units (True).

For curve E, we have:

Area E = (1/2 × 4 × 5) × 0.1

Area E = 10 × 0.1

Area E = 1 sq. units (True).

Read more on density curves here: brainly.com/question/26559908

#SPJ1

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Point E is the midpoint of AB and point F is the midpoint
GREYUIT [131]

The options are;

1) AB is bisected by CD

2) CD is bisected by AB

3) AE = 1/2 AB

4) EF = 1/2 ED

5) FD= EB

6) CE + EF = FD

Answer:

Options 1, 3 & 6 are correct

Step-by-step explanation:

We are told that Point E is the midpoint of AB. Thus, any line that passes through point E will bisect AB into two equal parts.

The only line passing through point E is line CD.

Thus, we can say that line AB is bisected by pine CD. - - - (1)

Also, since E is midpoint of Line AB, it means that;

AE = EB

Thus, AE = EB = ½AB - - - (2)

Also, we are told that F is the mid-point of CD.

Thus;

CF = FD

Point E lies between C and F.

Thus;

CE + EF = CF

Since CF =FD

Thus;

CE + EF = FD - - - (3)

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3 years ago
4 Tan A/1-Tan^4=Tan2A + Sin2A​
Eva8 [605]

tan(2<em>A</em>) + sin(2<em>A</em>) = sin(2<em>A</em>)/cos(2<em>A</em>) + sin(2<em>A</em>)

• rewrite tan = sin/cos

… = 1/cos(2<em>A</em>) (sin(2<em>A</em>) + sin(2<em>A</em>) cos(2<em>A</em>))

• expand the functions of 2<em>A</em> using the double angle identities

… = 2/(2 cos²(<em>A</em>) - 1) (sin(<em>A</em>) cos(<em>A</em>) + sin(<em>A</em>) cos(<em>A</em>) (cos²(<em>A</em>) - sin²(<em>A</em>)))

• factor out sin(<em>A</em>) cos(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (1 + cos²(<em>A</em>) - sin²(<em>A</em>))

• simplify the last factor using the Pythagorean identity, 1 - sin²(<em>A</em>) = cos²(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (2 cos²(<em>A</em>))

• rearrange terms in the product

… = 2 sin(<em>A</em>) cos(<em>A</em>) (2 cos²(<em>A</em>))/(2 cos²(<em>A</em>) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(<em>A</em>)

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - 1/cos²(<em>A</em>))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - sec²(<em>A</em>))

• divide through again by cos²(<em>A</em>)

… = (4 sin(<em>A</em>)/cos(<em>A</em>)) / (2/cos²(<em>A</em>) - sec²(<em>A</em>)/cos²(<em>A</em>))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(<em>A</em>) / (2 sec²(<em>A</em>) - sec⁴(<em>A</em>))

• factor out sec²(<em>A</em>) in the denominator

… = 4 tan(<em>A</em>) / (sec²(<em>A</em>) (2 - sec²(<em>A</em>)))

• rewrite using the Pythagorean identity, sec²(<em>A</em>) = 1 + tan²(<em>A</em>)

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (2 - (1 + tan²(<em>A</em>))))

• simplify

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (1 - tan²(<em>A</em>)))

• condense the denominator as the difference of squares

… = 4 tan(<em>A</em>) / (1 - tan⁴(<em>A</em>))

(Note that some of these steps are optional or can be done simultaneously)

7 0
3 years ago
A rectangular classroom has an area of 300 square feet. The length of the classroom is 5 feet longer than the width, w. Create a
weqwewe [10]

Answer:

w^2 + 5w = 300

Step-by-step explanation:

A= length x width

A =300

Width = w

length = w + 5

Therefore

300 = (w) x (w + 5)

300 = w^2 + 5w

w^2 + 5w = 300

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Answer:

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