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2 years ago

14

When constructing an inscribed square, how many lines will be drawn in the circle? (6 points) A. 2 B. 3 C .5 D. 7

1 answer:

nataly862011 [7]2 years ago

4 0

The answer is 2 or A.

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roger has a nail that is 12 centimeters long. he measures and records the length of the nail as 15 centimeters. what is the % er

Marianna [84]

15-12=3. 3/12=1/4=25% error

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2 years ago

Warren’s tape measure is marked in increments of 1/16 inch, not in decimal numbers. To measure the tiles, he needs to know the s

Bezzdna [24]

Answe

Step-by-step explanation:

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3 years ago

What is the value of x?<br><br> Enter your answer in the box. <br><br> x = ___

lapo4ka [179]

Answer:

x=8

Explanation:

The picture tells us that the triangle is equilateral (this means that each angle is 60°). We can assume this because each side has been marked with the same line.

Because each angle is equal to 60°, this means that:

7x+4=60

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7x=56
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3 years ago

7 times the sum of x and 5

kobusy [5.1K]

The answer is 7(x+5) ive had the exact question before!! Hope i helped u!!!!

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2 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