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3 years ago

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