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FromTheMoon [43]

3 years ago

6

1. On which road would there be a larger standard deviation in speed: a backroad or an interstate? Justify your answer

1 answer:

White raven [17]3 years ago

7 0

Answer:

Your answers may vary slightly. 5.2 Normal Distributions: Finding Probabilities If you are given that a random variable Xhas a normal distribution, nding probabilities corresponds to nding the area between the standard normal curve and the x-axis, using the table of z-scores. The mean (expected value) and standard deviation ˙should be given

Step-by-step explanation:

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The line shown below are parallel if the green line has a slope of -4 where is the slope of the redline

NemiM [27]

Answer:

-4

Step-by-step explanation:

Parallel lines have the same slope

If the green line has a slope of -4, then the red line has a slope of -4

4 0

3 years ago

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Rose picked 18 tulips and 20 daffodils. She divided the flowers into groups so that the same number of tulips and daffodils were

Naddika [18.5K]

Answer:

The greatest number of bouquets Rose could have is 2. 9 of each flower in each bouquet.

Step-by-step explanation:

5 0

3 years ago

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Evaluate 16^1/2 x 2^-3<br> Give your answer as a fraction in its simplest form.

kati45 [8]

Your answer would be 1/2

6 0

3 years ago

On a biased dice the probability of getting 1 is 0.3. The ice is rolled 150 times. How many ones would you expect to roll?

Montano1993 [528]

Answer: 45

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

Work Shown:

n = number of rolls = 150

p = probability of rolling a "1" = 0.3

n*p = expected number of "1"s that show up out of the 150 rolls

n*p = 150*0.3

n*p = 45

We expect about 45 ones will show up.

8 0

2 years ago

Find all real solutions and show work: <br> 2x^3-3x^2+18x-27=0

Amiraneli [1.4K]

Given:

The given equation is

$2x^3-3x^2+18x-27=0$

To find:

All the real solutions.

Solution:

We have,

$2x^3-3x^2+18x-27=0$

It can be written as

$x^2(2x-3)+9(2x-3)=0$

$(2x-3)(x^2+9)=0$

Using zero product property, we get

$(2x-3)=0$ and $(x^2+9)=0$

$2x=3$ and $x^2=-9$

$x=\dfrac{3}{2}$ and $x=\pm\sqrt{-9}$

We know that $x=\pm\sqrt{-9}$ is an imaginary number because there is a negative sign under the square root.

Therefore, $x=\dfrac{3}{2}$ is the only real solution of the given equation.

5 0

3 years ago