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2 years ago

6

A scatterplot is produced to compare the size of a lake to the number of fish that are in it. There are 15 data points, each rep

resenting a different lake. The points are widely dispersed on the scatterplot with no pattern of grouping. Interpret what the results of the scatterplot tell you about the relationship between the two variables.

2 answers:

topjm [15]2 years ago

8 0

There is no relation between the size of a lake and the number of fish in the lake.

yawa3891 [41]2 years ago

7 0

Answer:

Since there is no cluster formed in the scatterplot, the two variables are not related. Therefore, based on the data shown in the scatterplot, the number of fish in a lake is not dependent on the size of the lake.

Step-by-step explanation:

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Find the smallest zero for the function h(x) = 4x^2 - 8x - 60

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Sally is standing at one corner of a snowy rectangular field that measures 200 ft by 200 ft. She wishes toreach a warm cabin loc

pshichka [43]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The optimal point(i.e the position of P) that will allow Sally to reach the cabin as quickly as possible is $d = 50 \ ft$ from the cabin.

Step-by-step explanation:

From the question we are told that

The speed from the point Sally is standing to the point P is $u = 3 \ ft /s$

The speed from the point P to the cabin is $v = 5 \ ft /s$

Let denote the distance from the bottom left corner to the P be y ft

Generally according to Pythagoras theorem the distance from Sally's first position to point P is mathematically represented as

$s = \sqrt{y^2 + 200^2}$

Generally the distance from that point P to the cabin is mathematically represented as

$d = 200 -y$

Generally the time it takes Sally to cover that distance s is mathematically represented as

$t_1 = \frac{s}{u}$

=> $t_1 = \frac{ \sqrt{y^2 + 200^2}}{3}$

Generally the time it takes Sally to cover that distance d is mathematically represented as

$t_2 = \frac{d}{v}$

=> $t_2 = \frac{ 200 - y }{5}$

So the total time taken is

$t= \frac{ \sqrt{y^2 + 200^2}}{3} + \frac{ 200 - y }{5}$

Generally the minimum time taken with respect to the distance is mathematically evaluated by differentiating and equating the derivative to 0

So

$\frac{dt}{dy} =\frac{1}{3} \frac{y}{ \sqrt{y^2 + 200^2} } - \frac{1}{5} =0$

=> $\frac{dt}{dy} = \frac{y^2 }{y^2 + 200 ^2} =\frac{9}{25}$

=> $25y^2 = 9y^2 + 9 * 200^2$

=> $16y^2 = 360000$

=> $y = 150$

So the distance from point P to the cabin is

$d = 200 -150$

=> $d = 50 \ ft$

So the optimal point(i.e the position of P) that will allow Sally to reach the cabin as quickly as possible is $d = 50 \ ft$ from the cabin

8 0

1 year ago

Help please...<br> 15 points...one qustion<br> thank you

gavmur [86]

Hello!

Answer: $x=\frac{2}{3}$

Step-by-step explanation:

$7x-\frac{2}{3}=4$

Multiply by 3 from both sides of the equation.

$7x*3-\frac{2}{3}*3=4*3$

Refine.

$21x-2=12$

Add by 2 from both sides of the equation.

$21x-2+2=12+2$

Simplify.

$21x=14$

Divide by 21 from both sides of the equation.

$\frac{21x}{21}=\frac{14}{21}$

Simplify.

$x=\frac{2}{3}$

Hope this helps!

Thank you for posting your question at here on brainly.

-Charlie

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