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2 years ago

Match each set of vertices with the type of quadrilateral they form.

Mathematics

1 answer:

Serjik [45]2 years ago

5 0

Only correct the first one.

Please, see the graph in the attached file.

1) a rectangle with    A(3,3), B(3,6),
noncongruent ←→ C(7,6), D(7,3)
adjacent sides
(in green)

2) a parallelogram
with A(2,0), B(3,2)
nonperpendicular ←→ C(6,3), D(5,1)
adjacent sides
(in blue)

3) a square ←→ A(3,3), B(2,5),
(in red) C(4,6), D(5,4)

4) a rhombus with       A(2,-2), B(3,0),
nonperpendicular    ←→ C(4,-2), D(3,-4)
adjacent sides
(in black)

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What’s the answer? Because am having a hard time with this math problem.

Studentka2010 [4]

9514 1404 393

Answer:

34.5 square meters

Step-by-step explanation:

We assume you want to find the area of the shaded region. (The actual question is not visible here.)

The area of the triangle (including the rectangle) is given by the formula ...

A = 1/2bh

The figure shows the base of the triangle is 11 m, and the height is 1+5+3 = 9 m. So, the triangle area is ...

A = (1/2)(11 m)(9 m) = 49.5 m^2

The rectangle area is the product of its length and width:

A = LW

The figure shows the rectangle is 5 m high and 3 m wide, so its area is ...

A = (5 m)(3 m) = 15 m^2

The shaded area is the difference between the triangle area and the rectangle area:

shaded area = 49.5 m^2 - 15 m^2 = 34.5 m^2

The shaded region has an area of 34.5 square meters.

7 0

1 year ago

Kiersten runs a website that helps people learn programming. Every month, Kiersten receives a subscription fee of \$10$10dollar

bogdanovich [222]

Answer:

sorry but it is wrong the answer is 1,200

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Let ff be the function given by f(x)=e2x−1 x f(x)=e2x−1 x. Which of the following equations expresses the property that f(x)f(x)

navik [9.2K]

Answer:

Step-by-step explanation:

$\lim_{x \to \ 0} \frac{e^{2x}-1 }{x} \\= \lim_{x \to \0} \frac{e^{2x}-1 }{2 x} *2\\\rightarrow2 log e\\\rightarrow 2$

5 0

2 years ago

Read 2 more answers

Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum

dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

8 0

2 years ago

In estimating the mean score on a fitness exam, we use an original sample of size n = 30 and a bootstrap distribution containing

ioda

Answer:

C. 67.5 to 72.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.