A density curve for all the possible temperatures between 0°F and 40°F is a

If Pentagon ABCDE was dilated to create Pentagon A'B'C'DE', what rule was used?

svetlana [45]

Answer:

<em>B</em> $(x,y) \rightarrow \left(\frac{5}{2}x,\frac{5}{2}y\right)$

Step-by-step explanation:

<u>Dilations</u>

Given a point A(x,y) and a scale factor k the dilated image of A, called A' is calculated as A'=(kx,ky), assuming the same scale factor is applied in both axes.

The pentagon ABCDE was dilated to create pentagon A'B'C'D'E'. To find the dilaton rule used, we must find two clear points where the coordinates of both axes can be easily read from the graph.

Point C(-2,0) maps to C'(-5,0). This gives us the scale factor for the x-axis of -5/(-2)= 5/2.

The y-coordinate of E is 2 and the y-coordinate of E' is 5. This gives us the same scale factor for the y-axis of 5/2.

Thus, the rule to dilate the pentagon is:

B $\mathbf{(x,y) \rightarrow \left(\frac{5}{2}x,\frac{5}{2}y\right)}$

5 0

3 years ago

Empirical rule (statistics)Help please!!!

emmasim [6.3K]

The empirical rule you're referring to is the 68-95-99.7 rule, which asserts that for a normal (bell-shaped) distribution, approximately 68% of the distribution lies within 1 standard deviation of the mean; 95% lies within 2 standard deviations of the mean; and 99.7% lies within 3 standard deviations of the mean.

Let $X$ be the random variable denoting vehicle speeds along this highway. We want to find $P(61$ . To use the rule, we need to rephrase this probability in terms of the mean and standard deviation.

Notice that $61=70-9=70-3\cdot3$ , and $79=70+9=70+3\cdot3$ . In other words, 61 and 79 both lie exactly 3 standard deviations away from the mean, so $P(61$ .

8 0

3 years ago

Dora drove 80 km in 40 minutes then 120 km in 1 hour and then the final 200km took him 1 hour 20 minutes. what is his average sp

bonufazy [111]

9514 1404 393

Answer:

133 1/3 km/hour

Step-by-step explanation:

Average speed is computed as ...

average speed = (total distance)/(total time)

= (80 km +120 km +200 km)/(2/3 + 1 + 1 1/3 hour) = 400 km/(3 hour)

= 133 1/3 km/hour

8 0

3 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .

The area (in square inches) of the circle that the hexagon did not cover would be:

$\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}$ .

3 0

3 years ago

The odds of winning a rifle are 1:9 the probability of winning a carnival game is 0.15 does the raffle or the carnival game give

Lilit [14]

Answer:

Hence the carnival game gives you better chance of winning.

Step-by-step explanation:

Let the event of win be given by 1/10 in the game of rifle then the event of loose is given by 9/10

the

Odds in favor of a game are given by = P(Event)/ 1- P(Event)

Odds in favor of winning a rifle are given by = 1/10/ 1- 1/10

=1/10/9/10

=1/9

= 0.111

The probability of winning aa rifle game is 0.111

The probability of winning the carnival game is 0.15

Comparing the two probabilities 0.111:0.15

The probability of winning carnival game is greater than winning a rifle game

0.15>0.11

Hence the carnival game gives you better chance of winning.

3 0

3 years ago