A and B are two cylinders that are mathamatically similar.

Brrunno [24]

Answer:

50

Step-by-step explanation:

Each question is 6%/3 = 2% worth. If each question is worth 2% then total questions must be 50 (cuz that would be 2% x 50 (total question) = 100%).

4 0

2 years ago

4. a) A ping pong ball has a 75% rebound ratio. When you drop it from a height of k feet, it bounces and bounces endlessly. If t

Klio2033 [76]

First part of question:

Find the general term that represents the situation in terms of k.

The general term for geometric series is:

$a_{n}=a_{1}r^{n-1}$

$a_{1}$ = the first term of the series

$r$ = the geometric ratio

$a_{1}$ would represent the height at which the ball is first dropped. Therefore:

$a_{1} = k$

We also know that the ball has a rebound ratio of 75%, meaning that the ball only bounces 75% of its original height every time it bounces. This appears to be our geometric ratio. Therefore:

$r=\frac{3}{4}$

Our general term would be:

$a_{n}=a_{1}r^{n-1}$

$a_{n}=k(\frac{3}{4}) ^{n-1}$

Second part of question:

If the ball dropped from a height of 235ft, determine the highest height achieved by the ball after six bounces.

$k$ represents the initial height:

$k = 235\ ft$

$n$ represents the number of times the ball bounces:

$n = 6$

Plugging this back into our general term of the geometric series:

$a_{n}=k(\frac{3}{4}) ^{n-1}$

$a_{n}=235(\frac{3}{4}) ^{6-1}$

$a_{n}=235(\frac{3}{4}) ^{5}$

$a_{n}=55.8\ ft$

$a_{n}$ represents the highest height of the ball after 6 bounces.

Third part of question:

If the ball dropped from a height of 235ft, find the total distance traveled by the ball when it strikes the ground for the 12th time. 

This would be easier to solve if we have a general term for the <em>sum </em>of a geometric series, which is:

$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$

We already know these variables:

$a_{1}= k = 235\ ft$

$r=\frac{3}{4}$

$n = 12$

Therefore:

$S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{1-\frac{3}{4} }$

$S_{n}=\frac{(235)(1-\frac{3}{4} ^{12})}{\frac{1}{4} }$

$S_{n}=(4)(235)(1-\frac{3}{4} ^{12})$

$S_{n}=910.22\ ft$

8 0

3 years ago

1. Solve triangle ABC if A = 42.3º, b = 12.9 meters, and c = 15.4 meters.

kumpel [21]

Answer:

a = 10.36m

B = 47.7º

C = 90º

Step-by-step explanation:

we know the angle of A so we do the sin of A; Spoiler: its 0.67

0.67 = a / 15.4

0.67 * 15.4 = a

a = 10.36

----------

B = 90 - 42.3 = 47.7º

6 0

3 years ago

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

Please answer 20 questions i WILL MARK YOU BRAINLIEST PLEASE HELP ME 5 QUESTIONSSSSS PLEASEEE HELPPP

allochka39001 [22]

The answers should be 1D, 2C, 3B, 4A, and 5A.

8 0

3 years ago

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