Ask question

Ask question

All categories

2 years ago

15

A sports announcer states that there are 65,000 fans in the stadium. Estimate the maximum noise level. Is this estimate reasonab

le? Explain your reasoning.

1 answer:

SIZIF [17.4K]2 years ago

8 0

Question:

The scatter plot shows the maximum noise level when different numbers of people are in a stadium. The linear model is given by the equation y=1.5x + 22.7 where y represents maximum noise level and x represents the number of people, in thousands, in the stadium.

A sports announcer states that there are 65,000 fans in the stadium. Estimate the maximum noise level. Is this estimate reasonable? Explain your reasoning.

Answer:

$y = 97522.7$

Step-by-step explanation:

Given

$y = 1.5x + 22.7$

$x = 65000$

Required

Find y

Substitute 65000 for x in the given equation

$y = 1.5 * 65000 + 22.7$

$y = 97500 + 22.7$

$y = 97522.7$

<em>It is reasonable because as the population increases, the noise level also increases</em>

You might be interested in

Vector v is shown in the graph.

Elis [28]

Question 4

<h3>Answer: Choice D</h3>

Explanation:

If the initial point is the origin, the coordinates of the terminal point form the vector itself in component form. We go from (-8,6) to <-8,6>. The notation change is from ordered pair to vector format.

We have a right triangle with legs of 8 and 6 units. The pythagorean theorem will help us determine the hypotenuse is 10. Therefore, the vector length is 10 units and we would say ||v|| = 10. We have a 6-8-10 right triangle.

===========================================================

Question 5

<h3>Answer: Choice B</h3>

Explanation:

Vector v starts at (2,5) and ends at (-3,-2).

The x component of the vector is x2-x1 = -3-2 = -5 meaning we move 5 units to the left when going from the start point to the endpoint.

At the same time we move 7 units down because y2-y1 = -2-5 = -7 which is the y component of the vector.

The component form of vector v is

v = <-5, -7>

it says "move 5 units left, 7 units down".

Apply the pythagorean theorem to find the length of the vector.

a^2+b^2 = c^2

c = sqrt(a^2 + b^2)

||v|| = sqrt( (-5)^2 + (-7)^2 )

||v|| = 8.602 which is approximate.

Now let's use the arctangent function to find the angle

theta = arctan(b/a)

theta = arctan(-7/(-5))

theta = 54.462 which is also approximate.

There's a problem however. This angle is in Q1 but the vector <-5,-7> is in Q3. An easy fix is to add on 180 to rotate to the proper quadrant.

54.462+180 = 234.462

which is the proper approximate angle for theta.

6 0

1 year ago

3. What is the slope of a line that passes<br> through (0,-5) and (8,27)?

Elden [556K]

Answer:

4.

Step-by-step explanation:

Slope = difference in y coordinates / difference in the corresponding x coordinates

(27 - (-5)) / (8 - 0)

= 32 / 8

= 4.

6 0

2 years ago

Read 2 more answers

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

7 0

2 years ago

Pls I’m in the need of help right now

padilas [110]

25 girls and 50 boys!!

49 divided by 7 = 7.

So there were 7 adults.

49 + the 1 it mentioned = 50 boys

50 divided in half = 25 girls.

50+25+7 = 82.

4 0

2 years ago

Solve: |4x + 5| + 8 >(or equal to)45

KatRina [158]

This is pretty simple once you get used to this

First set it equal to the original and the negative one
(>_) is the bigger and equal to part

4x+5>_45 or 4x+5>_-45

Then solve for x for both

4x>_40. 4x>_-50
X>_10 X>_-25/2

8 0

3 years ago