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3 years ago

5

1. A sample is being chosen for a survey to find out how people feel about adding more stop signs near a school. Which group wou

ld be least biased?
A. voters
B. drivers
C. students
D. teachers

1 answer:

Anna71 [15]3 years ago

8 0

I think that it would be voters because they are asking the people if they should or not.

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-15= x/-0.5 8= what does x equal

Stels [109]

8*0.5*x=-15
x = -15/4

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3 years ago

GREYUIT [131]

Answer:

Step-by-step explanation:

1.

James goes to an arcade.

He has one go on the Teddy Grabber.

He has one go on the Penny Drop.

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The probability that he wins on the Penny Drop is 0.3.

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3 years ago

A child who is 4 ½ feet tall casts a 6-foot shadow. At the same time, a nearby statue casts a 15-foot shadow. How tall is the st

stiv31 [10]

Answer:

13 1/2 tall

Step-by-step explanation:

If 4 1/2 is the childs hight and their shadow is 6 feet, the shadow adds 2 1/2 to the height.

5 0

2 years ago

Is 4/12more or less than 4/6

11111nata11111 [884]

4/12 is less than 4/6

4/12 = 2/6

and 2/6 < 4/6

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3 years ago

Read 2 more answers

An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i

Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = $\sqrt{(15-x)^2+1^2}$

The total time, 't', is given as follows;

$t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}$

The minimum value of 't' is found by differentiating with an online tool, as follows;

$\dfrac{dt}{dx} = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} = \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }$

At the maximum/minimum point, we have;

$\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0$

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

$t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89$

Therefore, the distance to run, x ≈ 14.96 mile

6 0

2 years ago