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Yakvenalex [24]
3 years ago
13

Doug makes a parallelogram-shape sign for his shop.

Mathematics
2 answers:
ser-zykov [4K]3 years ago
6 0

Answer:

C, 14.28

Step-by-step explanation:

area = base * height

= (1.9+3.2) * 2.8

=14.28

swat323 years ago
3 0

Answer:

Answer is C! brainliest this time pls

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Pls answer quickly. Thanks!
Gwar [14]

Answer:

the answer is D

Step-by-step explanation:


4 0
3 years ago
if someone could please please help me with this I would appreciate it , if you could also explain your answer that would be gre
weqwewe [10]

Answer:

height for the building = 551.8 ft

Step-by-step explanation:

The info given describes the shape of a right triangle.

Thus:

Reference angle = Angle of elevation = 23°

Opposite side = height of the building = ?

Adjacent side = 1,300 ft

Apply trigonometric function, TOA:

Tan 23 = Opp/Adj

Tan 23 = Opp/1,300

1,300 × Tan 23 = Opp

551.817261 = Opp

Opp ≈ 551.8 ft

height for the building = 551.8 ft

3 0
3 years ago
The volume of a rectangular prism is 3,240 ft3. The prism is 9ft long and 12ft wide. How tall is the rectangular prism? Please h
Softa [21]
3.5*10^5
''''''''''''''''
5 0
3 years ago
If you walked 2 3/4 miles on Monday and then walked 3 1/3 miles on Tuesday, how many miles did you walk during those two days?
Anit [1.1K]
6 miles that is rounded up
8 0
3 years ago
Read 2 more answers
The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for
Blababa [14]

f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}

a. 9:00 AM is the 60 minute mark:

f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982

The probability of doing so for at least 2 of 5 days is

\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x

Then the desired probability is

F_X(30)-F_X(15)\approx0.950-0.777=0.173

7 0
3 years ago
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