What 50 precent of 7

SHOW YOU WORK AND EXPLAIN PLS I WILL MARK YOU BRAINLIEST PLEASE.

KatRina [158]

Answer:

a) 1.8 $cm^2$

b) 3.6 $feet^2$

Step-by-step explanation:

Conversion and calculation of areas

The area of a rectangle is A=wh, where w is the width and h is the height. There are 3 feet in one yard

a)

We are told Kalie put 6 stickers, each one of 1/2 centimeters (0.5 cm) wide by 3/5 (0.6 cm) centimeter long. The area of one sticker is

A=(0.5)(0.6)=0.3 $cm^2$

Assuming there is no overlapping, the 6 stickers have a total area

6*0.3 $cm^2$ =1.8 $cm^2$

b)

Each of Elana's wrapping papers measures 2/5 yards long and 1/4 yard wide. Converting them to feet we have

long=2/5*3=1.2 feet

wide=1/4*3=0.75 feet

Area of each paper=1.2 feet*0.75 feet=0.9 $feet^2$

Area of the entire board, assuming no overlapping and no space left uncovered=4*0.9 $feet^2$

Area of board=3.6 $feet^2$

7 0

3 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

4 0

3 years ago

Let AA and BB be events such that P(A∩B)=1/73P(A∩B)=1/73, P(A~)=68/73P(A~)=68/73, P(B)=21/73P(B)=21/73. What is the probability

krok68 [10]

Answer:

$P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}$

Step-by-step explanation:

Let A and B events. We have defined the probabilities for some events:

$P(A') =\frac{68}{73} , P(B) =\frac{21}{73} , P(A \cap B) =\frac{1}{73}$

Where A' represent the complement for the event A

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$