Mathematics High School

On a day when the speed of sound in air is 94.0 km/hr you yell across a canyon. you hear your voice echo back from the opposite canyon wall 1.03 min after you first yell. what is your distance in meters from the other side of the canyon?
Whoever does this thank you so much, you're doing God's work!

Answers

Answer 1

Answer:

94.0 km/hr * 1.03 min = 96.82 km/m

Answer 2

Answer:

1,613.7 m

Step-by-step explanation:

speed of sound in air = 94 km/hr.

with a time of 1.03 min.

find distance traveled in meters

solution:

distance = velocity x time

= 94 km x 1 hr. x 1.03 min x 1 000 m

hr. 60 min. 1 km

= 1,613.7 m

Related Questions

The age of 10 factory workers is shown in the table. If a 58 year-old worker is added, by how much will this increase the median age of the workers?

Answers

Answer: 3 years

Step-by-step explanation:

usa test prep

The F-ratio increases as _________. the variability between means increases relative to the variability within groups the variability between means decreases relative to the variability within groups the total variability increases the total variability decreases

Answers

Answer: variability between means increases relative to the variability within groups

Step-by-step explanation:

The formula to calculate F-ratio : variance between groups divided by variance within groups

Here, F-ratio $\propto$ is proportional to variance between groups but inversely proportional to variance within groups .

So, F-ratio increases when variability between means increases relative to the variability within groups.

Method: Forty-eight preschoolers (age range =3 years 6 months to 4 years 5 months; 24 boys, 24 girls) watched a video featuring two female actors seated side by side. On each of four trials, one actor announced her intention to hide a ball under one of four cups; the other agreed, covered her eyes, and turned around to face the back wall. The hider placed a small barrier in front of the cups so that the children could watch as she hid the ball but could not see the particular cup she baited. She announced that she had finished and removed the barrier. Both actors faced the camera throughout the rest of the trial. The children were randomly assigned to three conditions. In the point condition (n=16; mean age =3 years 11 months), the two actors simultaneously pointed to different cups. In the grasp condition ( n=16; mean age =4 years 0 months), they simultaneously grasped the tops of different cups. We used grasping as a comparison gesture because young children understand that it, like pointing, is intentional and object-directed (Woodward, 1999). However, it is not often used communicatively, and so may not be as likely as pointing to lead to the misattribution of knowledge. In the baseline condition ( n=16; mean age =3 years 11 months), the two actors simply sat with their hands in their laps. After the actors gestured (or not), the experimenter paused the video and asked, "Who knows where the ball is?" One actor hid the ball on the first and fourth trials; the other actor did so on the middle two trials. Results: Children in the grasp and baseline conditions selected the actor who hid the ball as the one who knew its location more frequently than would be expected by chance, ts (15)>3.74, ps <.01,ds>0.94; specifically, children in the grasp location selected that actor on 3.13(SD=1.20) of the 4 trials, and children in the baseline condition selected her on 3.25(SD=1.06) of the 4 trials. Children in the point condition performed at chance level, t(15)<1, selecting the hider on just 2.13(SD=1.25) trials. An analysis of variance revealed a significant effect of condition, E(2,45)=4.38,p=.018,η2=.16. Children in the point condition were less likely to select the hider than were children in the grasp or baseline condition (Tukey's honestly significant difference, ps<.05 ); performance did not differ between the latter two conditions. Seeing the two actors point may have led children in the point condition to assume that both were knowledgeable. But it is also possible that these children ignored the test question: Rather than indicating which pointer was knowledgeable, perhaps they reflexively indicated where they would search for the ball, which would also lead to chance performance (Palmquist, Burns, \& Jaswal, 2012; Povinelli \& deBlois, 1992). One reason to doubt this possibility is that the children tended to respond to the test question by pointing to an actor's face rather than one of the cups (72\% vs. 28% of trials). However, to investigate this possibility directly, we conducted a control study with 8 different children (mean age =3 years 9 months, range =3 years 6 months to 4 years 1 month; 5 boys, 3 girls). The procedure was the same as in the point condition, but after the actors pointed, the experimenter asked, "Who hid the ball?" If pointing automatically triggers a search response, even when the test question does not ask children to indicate where they would search, they should select the two pointers equally (as they did in the point condition) when this procedure is followed. In fact, however, the children correctly indicated the hider on 3.89(SD=0.35) of the 4 trials, more often than expected by chance, t(7)=15.00,p<.001,d=7.59. Thus, children do not automatically respond to pointing by

Answers

The study found that preschoolers in the grasp and baseline conditions recognized the actor who hid the ball as the one with knowledge of its location, suggesting that pointing gestures influenced their judgments.

In this study, 48 preschoolers participated, ranging in age from 3 years 6 months to 4 years 5 months, with an equal distribution of 24 boys and 24 girls. The children watched a video featuring two female actors seated side by side.

The actors engaged in a task where they hid a ball under one of four cups, while the other actor covered her eyes and turned around. A small barrier was placed in front of the cups, preventing the children from seeing the specific cup where the ball was hidden.

In the grasp condition, the actors simultaneously grasped the tops of different cups.

The baseline condition served as a comparison, where the actors simply sat with their hands in their laps. After the actors performed the gestures or remained in the baseline condition, the video was paused, and the children were asked, "Who knows where the ball is?"

The results showed that children in the grasp and baseline conditions selected the actor who hid the ball as the one who knew its location more frequently than would be expected by chance.

In contrast, children in the point condition performed at chance level, indicating the hider on just 2.13 out of 4 trials

An analysis of variance revealed a significant effect of condition, suggesting that the pointing gesture influenced the children's judgments.

The possibility that children in the point condition ignored the test question and instead reflexively indicated where they would search for the ball was considered.

The results showed that children correctly indicated the hider more often than expected by chance, indicating that they were not simply responding to the pointing gesture.

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can yall help me solve this (x + 8)° + 4x° the subject is actually geometry

Answers

Answer:

Step-by-step explanation:

If a1 = 7 and an = an-1 + 4 then find the value of a4.

Answers

Answer:

$a_4=19$

Step-by-step explanation:

Given:

$a_1=7$$a_n=a_{n-1}+4$

$\implies a_2=a_{1}+4=7+4=11$

$\implies a_3=a_{2}+4=11+4=15$

$\implies a_4=a_{3}+4=15+4=19$

Put values get answer

a_n=a_(n-1)+4a_4=a_3+4a_4=a_2+4+4a_4=a_1+4+4+4a_4=7+12a_4=19

Carla has a rectangular garden in her backyard. The width of the garden is 9 meters. The area of the garden is 360 square meters. What is the length of the garden? Show your work.

Answers

Answer:

l = 40 m

Step-by-step explanation:

The formula for area of a rectangle is

A = lw, where

A is the area in square units,l is the length,and w is the width

Step 1: Since we're given the area and width, we plug in these two values for A and w and to solve for l (length):

360 = 9l

Step 2: Divide both sides by 9 to solve for l

(360 = 9l) / 9

40 m = l

Optional Step 3: We can check our answers by checking that the product of 40 and 9 is 360

40 * 9 = 360

360 = 360

what do 20-(12+6) equal

Answers

Answer:

Step-by-step explanation:

Since the 12 + 6 are parenthesis, we do these first

12 + 6 = 18

Then will solve the rest, 20 - 18

20 - 18 = 2, therefore leaving us with the answer of 2

Slope Please help!!!!!

Answers

Answer:

2/3

Step-by-step explanation:

2/3

Step-by-step explanation:

Very Simple

Take 2 points

x,y

x1,y1= 3,2

x2,y2=6,4

Find the slope

Y2-Y1/X2-X1

4-2/6-3

2/3

(ANSWER NOW FOR 20 POINTS AND BRAINLEST) Quadrilateral EFGH is a scaled copy of quadrilateral ABCD. Answer the statement below as true or false. Segment EF is twice as long as segment AB.

True or False?

Answers

Answer:true

Step-by-step explanation:

The top line segment (12) is twice as long as the original, smaller (6), segment

An icicle that is 12 mm long melts at a constant rate of 1.2 mm per minute. Let t represent the number of minutes since the icicle started melthing and suppose R is a function such that R(t) represents the remaining length of the icicle (in mm) t minutes after it started melting.
A.Write a function formula for R in terms of t.
b.What is the domain of R relative to this context? Enter your answer as an interval.
c.What is the range of R relative to this context? Enter your answer as an interval.
d.What is the domain of R−1 relative to this context? Enter your answer as an interval.
e.What is the range of R−1 relative to this context? Enter your answer as an interval.
f.Solve R(t)=6.6 for t.
t= g.What does your solution in part (f) represent in this context? Select all that apply.
How long it takes for the icicle to melt completely.
How many minutes since the icicle started melting when it is 6.6 mm long.
The length of the icicle (in mm) 6.6 minutes after it started melting.

Answers

R(t)=6.6 for t, 5 minutes, the domain of R relative to this context [0, 65 ÷ 6], the range of R related to this context, the domain of R relative to this context, [0, 12], and the range of R relative to this context.

The values of variables, domain, and range:

a) 12 - 1.2t, Using the build equation.

b) Least length = 0 mm.

max time = 12 - 1.2t = 0

t = 65 ÷ 6 minutes.

domain t ∈ [0, 65 ÷ 6]

c) range, R ∈ [0, 12]

d) domain of $R^{-1}$ [0, 12]

e) domain of $R^{-1}$ [0, 65 ÷ 6]

f) 12 - 1.2t = 7

1.2t = 7

t = 5

R(t)=6.6 for t, 5 minutes.

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Directions: Use the ruler to measure the line segments.

Answers

The length of each line a , b and C are 0.1875, 0.5625 and 1 inch(es) respectively

From the measuring rule given ;

Each successive tick marks is (1/16) = 0.0625 inches

Therefore, using the value per tick value calculated above , we can deduce the length of the each line.

The measure of 'a':

3 ticks * 0.0625 = 0.1875 inches

The measure of 'b':

9 ticks * 0.0625 = 0.5625 inches

The measure of 'c':

16 ticks * 0.0625 = 1 inch

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Find the values of c such that the area of the region bounded by the parabolas

y = 4x^2 − c^2 and y = c^2 − 4x^2

is 36. (Enter your answers as a comma-separated list.)

Answers

Here's the formula written in LaTeX code:

The value of $\(c$\) that satisfies the given condition is -3. To find the values of $\(c$\) such that the area of the region bounded by the parabola $\(y = 4x^2 - c^2$ and $y = c^2 - 4x^2$\) is 36, we need to set up the integral to find the area between the two curves and then solve for $c$.

The area between two curves can be found by integrating the difference between the upper and lower curves with respect to $\(x$\).

First, let's set the two equations equal to each other to find the $\(x$\)-values where the curves intersect:

$\[4x^2 - c^2 = c^2 - 4x^2.\]$

Simplifying this equation, we get:

$\[8x^2 = 2c^2.\]$

$\[x^2 = \frac{c^2}{4}.\]$

Taking the square root of both sides, we get:

$\[x = \pm \frac{c}{2}.\]$

Now, let's set up the integral to find the area:

$\[\text{{Area}} = \int_{x_1}^{x_2} [f(x) - g(x)] dx,\]$

where $\(x_1$ and $x_2$ are the $x$-values where the curves intersect, $f(x)$ is the upper curve ($4x^2 - c^2$), and $g(x)$ is the lower curve ($c^2 - 4x^2$).\)

Using the $\(x$\)-values we found earlier, the integral becomes:

$\[\text{{Area}} = \int_{-\frac{c}{2}}^{\frac{c}{2}} [(4x^2 - c^2) - (c^2 - 4x^2)] dx.\]$

Simplifying this expression, we get:

$\[\text{{Area}} = \int_{-\frac{c}{2}}^{\frac{c}{2}} (8x^2 - 2c^2) dx.\]$

Integrating, we get:

$\[\text{{Area}} = \left[\frac{8}{3}x^3 - 2c^2x\right]\Bigg|_{-\frac{c}{2}}^{\frac{c}{2}}.\]$

Evaluating this expression at the limits of integration, we get:

$\[\text{{Area}} = \left[\frac{8}{3}\left(\frac{c}{2}\right)^3 - 2c^2 \left(\frac{c}{2}\right)\right] - \left[\frac{8}{3}\left(-\frac{c}{2}\right)^3 - 2c^2 \left(-\frac{c}{2}\right)\right].\]$

Simplifying further, we get:

$\[\text{{Area}} = \frac{2c^3}{3} - c^3 + \frac{2c^3}{3} - c^3.\]$

$\[\text{{Area}} = \frac{4c^3}{3} - 2c^3.\]$

Now, we can set this expression equal to 36 and solve for $\(c$\) :

$\[\frac{4c^3}{3} - 2c^3 = 36.\]$

Multiplying through by 3 to clear the fraction, we get:

$\[4c^3 - 6c^3 = 108.\]$

Simplifying further, we get:

$\[-2c^3 = 108.\]$

Dividing by -2, we get:

$\[c^3 = -54.\]$

Taking the cube root of both sides, we get:

$\[c = -3.\]$

Therefore, the value of $\(c$\) that satisfies the given condition is -3.

In summary, the value of $\(c$\) such that the area of the region bounded by the parabolas $\(y = 4x^2 - c^2$\) and $\(y = c^2 - 4x^2$\) is 36 is $\(c = -3$.\)

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In ΔCDE, the measure of ∠E=90°, CD = 9.2 feet, and DE = 8.3 feet. Find the measure of ∠C to the nearest tenth of a degree.

pls help!!

In CDE, the measure of E=90, CD = 9.2 feet, and DE = 8.3 feet. Find the measure of C to the nearest tenth

Answers

Answer:

I need the answer to plz

Step-by-step explanation:

i don't know either sorry

Answer:

64.4

Step-by-step explanation:

deltamath gave me the answer

what is the probability that a person selected randomly from the survey group does not have a two-handed backhand swing, given that the person is male, or p(no two-handed backhand/male)?

Answers

The probability that a person selected randomly from the survey group does not have a two-handed backhand swing, given that the person is male is given as follows:

P(no two-handed backhand/male) = 0.81 = 81%.

How to obtain a probability?

The probability of an event in the context of an experiment is obtained as the number of desired outcomes of the experiment divided by the number of total outcomes of the experiment.

In the context of this problem, these amounts are given as follows:

Desired: No two-handed backhanded swing and male.Total: Male.

The number of males that have a two-handed backhand is of:

256 - 176 = 80.

Hence the number that does not have is of:

421 - 80 = 341.

Then the probability is of:

p = 341/421 = 0.81 = 81%.

Missing Information

The problem is given by the image at the end of the answer.

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what is the probability that a person selected randomly from the survey group does not have a two-handed

Estimate and then solve 1,596 ÷ 42 = ___. (4 points)

a
380

b
400

c
38

d
40

Answers

Answer:

Step-by-step explanation:

Estimating 1,596 will give you 1,600. Estimating 42 will give you 40. Dividing 1,600 ÷ 40 will give you 40. So, the answer would be D.

Find X.
Math it’s.com

Answers

Answer:

value of x = 8

Step-by-step explanation:

well here,

line x is parallel to the line measuring 18

so the two triangles are similar to each other

hence ratio of their sides are equal

so 6/ 12 = x/ 16

=》1/2 = x/16

=》 x = 16 / 2

=》 x = 8

i hope u got it

have a nice day

Answer:

x = $\frac{16}{3}$

Step-by-step explanation:

The outer and inner triangles are similar, thus the ratios of corresponding sides are equal, that is

$\frac{6}{6+12}$ = $\frac{x}{16}$ , that is

$\frac{6}{18}$ = $\frac{x}{16}$ ( cross- multiply )

18x = 96 ( divide both sides by 18 )

x = $\frac{96}{18}$ = $\frac{16}{3}$

two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 2t, 1 6t, 1 14t . find the points at which their paths intersect. (if an answer does not exist, enter dne.

Answers

The answer is: (0, 0, 0).
To find the points at which the paths of the two particles intersect, we need to find the values of t for which r1(t) = r2(t). Equating the x, y, and z coordinates, we get three equations:

t = 1/2t
t2 = 1/6t
t3 = 1/14t
Simplifying the first equation, we get t = 0, which means the particles intersect at the origin. Substituting t = 0 into the second and third equations, we get 0 = 0 and 0 = 0, respectively. This confirms that the particles do indeed intersect at the origin. Therefore, the answer is: (0, 0, 0).

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find the value of each trigonometric ratio

Answers

The trigonometric relations from the triangles are

a) tan A = 5/12

b) sin C = 3/5

c) cos X = 3/5

d) sin Z = 4/5

e) tan Z = 4/3

f) tan X = 12/5

What are trigonometric relations?

Trigonometry is the study of the relationships between the angles and the lengths of the sides of triangles

The six trigonometric functions are sin , cos , tan , cosec , sec and cot

Let the angle be θ , such that

sin θ = opposite / hypotenuse

cos θ = adjacent / hypotenuse

tan θ = opposite / adjacent

tan θ = sin θ / cos θ

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

Given data ,

The triangle is ΔABC

tan A = opposite side / adjacent side