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2 years ago

Pls help me :(

Mathematics

1 answer:

Ann [662]2 years ago

3 0

Answer:

Step-by-step explanation:

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What is the approximate area of a circle with a diameter of 28 centimeters? Use π = 3.14.

larisa [96]

Formula is π * r^2

π = about 3.14 and the radius is half of the diameter so 28/2 = 14

3.14 * 14^2 = 615.44

B

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3 years ago

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Given that R=9x+8y find x when y=8 and R=4

Ilya [14]

Answer:

Step-by-step explanation:

R = 9x + 8y

4 = 9x + 8*8

4 = 9x + 64

Subtract 64 from both sides

4 - 64 = 9x + 64 - 64

-60 = 9x

Divide both sides by 9

-60/9 = 9x/9

-20/3 = x

x = -6 2/3

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3 years ago

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I rlly need help on this guys it’s due tomorrow!! The outside radius of a truck tire is 11 inches. Approximately how far will th

Sergeeva-Olga [200]

Answer:

207.24 inches

Hope this helps

Step-by-step explanation:

first you need to find the circumference...

2pi R (r=radius=11)

2 x 3.14 x 11 = 69.08 (i'm using 3.14 as pi so yea...)

3 rotations...

69.08 x 3= 207.24 inches

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3 years ago

The cost to play the game is $2. The winner gets $6 for making green. Suppose 36 people play the game. How much money will the s

IRINA_888 [86]

Answer:

$72

Step-by-step explanation:

I did 36 x 2. I did not subtract the -6 for the prize money because it says not to.

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3 years ago

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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