What did the abacus say when asked if he could help out at the school carnival worksheet answer

PLEASE HELP ASAP: A particle is moving with velocity v(t) = t2 – 9t + 18 with distance, s measured in meters, left or right of z

dimaraw [331]

1) $\frac{v(8)-v(0)}{8 - 0} = \frac{10-18}{8} = -1\frac{m}{s}$ .
2) $v(5) = 5^2-9*5+18 = 25-45+18 = -2\frac{m}{s}$ .
3) The particle is moving right when the velocity function is positive: $0\ \textless \ t\ \textless \ 3$ or $6\ \textless \ t\ \textless \ 8$ .
4) When $0\ \textless \ t\ \textless \ \frac{9}{2}$ the particle is slowing down because the acceleration is close to zero $\Rightarrow$ the particle is speeding up when acceleration is increasing away from zero: $\frac{9}{2}\ \textless \ t\ \textless \ 8$ .
5) $(\frac{1}{8})* \int\limits^8_0 {t^2-9t+18dt}=\frac{1}{8}*(\frac{t^3}{3}-(\frac{9}{2}*t^2+18t)_{0}^{8}= \\=(\frac{1}{8})*(\frac{8^3}{3}-(\frac{9}{2})*8^2+18*8)=\frac{8^2}{3}-(\frac{9}{2})*8+18=3\frac{1}{3} \frac{m}{s}$ .

3 0

2 years ago

You and a group of friends are going to a five-day outdoor music festival during spring break. You hope it does not rain during

KIM [24]

Answer:

The probability of not rain during the entire festival is 0.19

Step-by-step explanation:

The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1.

In this case we have the probability of raining of each day, we need the probability of NOT raining.

First day= 45% chance of rain, the complement is 55%

Second day 55% chance of rain, the complement is 45%

Third day a 10% chance of rain, the complement is 90%

Fourth day a 10% chance of rain, the complement is 90%

Fifth day 5% chance of rain, the complement is 95%

To get the probability of NOT raining in the entire festival is the multiplication of all the complements.

P(not raining) = 0.55 x 0.45 x 0.90 x 0.90 x 0.95= 0.19

4 0

2 years ago

A rectangular yard is fenced in using 319 feet of custom fence Your friends really like the fence and decide to fence in their y

Vsevolod [243]

We have been given that a rectangular yard is fenced in using 319 feet of custom fence Your friends really like the fence and decide to fence in their yard using the same fence. Their yard is similar but has a scale factor of 8/5 times the size of yours.

We are asked to find the length of fence that your friends should purchase.

Since the size of your friend's yard is $\frac{8}{5}$ times your yard, so your friend should purchase $\frac{8}{5}$ times the 319 feet.