Which of the following is the definition of a hypothesis?

Please help me! I think this 20 character thing is unnecessary when you have an attachment

PIT_PIT [208]

Answer:

B

Step-by-step explanation:

Also the answer must be under 9 so that rules out option D

Because We know CE = 2 and AD is visibly larger than CE, 6We can rule out option A

And that brings us down to B and C

CB is equal to 6, And I'm no genius, but I'm pretty sure AD is DEFINATLY not equal to 6.

so the answer is B

8 0

3 years ago

Factor by grouping.

hram777 [196]

The correct answers are A C and B

4 0

3 years ago

Read 2 more answers

Recall that in a 30 – 60 – 90 triangle, if the shortest leg measures x units, then the longer leg measures xStartRoot 3 EndRoot

Brums [2.3K]

The area of the shaded region is $\rm (150\sqrt{3} \ - 75\pi ) \ feet^2$ option first is correct.

It is given that a circle is inscribed in a regular hexagon with sides of 10 feet.

It is required to find the shaded area (missing data is attached shown in the picture).

<h3>What is a circle?</h3>

It is defined as the combination of points that and every point has an equal distance from a fixed point ( called the center of a circle).

We have a hexagon with a side length of 10 feet.

We know the area of the hexagon is given by:

$\rm A = \frac{3\sqrt{3} }{2} a^2$ where a is the side length.

$\rm A = \frac{3\sqrt{3} }{2} 10^2$ ⇒ $150\sqrt{3}$ $\rm feet^2$

We have the shortest length = x feet and from the figure:

2x = 10

x = 5 feet

The radius of the circle r = longer leg

$\rm r = x\sqrt{3} \Rightarrow 5\sqrt{3}$ feet

The area of the circle a = $\pi r^2$ ⇒ $\pi (5\sqrt{3} )^2 \Rightarrow 75\pi \ \rm feet^2$

The area of the shaded region = A - a

$\rm =(150\sqrt{3} \ - 75\pi ) \ feet^2$

Thus, the area of the shaded region is $\rm (150\sqrt{3} \ - 75\pi ) \ feet^2$

Learn more about circle here:

brainly.com/question/11833983

7 0

2 years ago

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla

svp [43]

Here is the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29: 117 122 129 118 131 123

Men aged 60-69: 130 153 141 125 164 139

Group of answer choices

a)

Men aged 20-29: 4.8%

Men aged 60-69: 10.6%

There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

$coefficient \ of \ variation = \dfrac{standard \ deviation}{mean}*100$

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = $\dfrac{\sum \limits^{n}_{i-1}x_i}{n}$

Mean = $\frac{117+122+129+118+131+123}{6}$

Mean = $\dfrac{740}{6}$

Mean = 123.33

Standard deviation $= \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }$

Standard deviation = $\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }$

Standard deviation = $\sqrt{\dfrac{161.3334}{5}}$

Standard deviation = $\sqrt{32.2667}$

Standard deviation = 5.68

The $coefficient \ of \ variation = \dfrac{standard \ deviation}{mean}*100$

$coefficient \ of \ variation = \dfrac{5.68}{123.33}*100$

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = $\dfrac{\sum \limits^{n}_{i-1}x_i}{n}$

Mean = $\frac{ 130 + 153 + 141 + 125 + 164 + 139}{6}$

Mean = $\dfrac{852}{6}$

Mean = 142

Standard deviation $= \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }$

Standard deviation = $\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }$

Standard deviation = $\sqrt{\dfrac{1048}{5}}$

Standard deviation = $\sqrt{209.6}$

Standard deviation = 14.48

The $coefficient \ of \ variation = \dfrac{standard \ deviation}{mean}*100$

$coefficient \ of \ variation = \dfrac{14.48}{142}*100$

Coefficient of variation = 10.2% for men age 60 - 69

Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

4 0

3 years ago

Alex has 4 less than twice the number baseball cards as leo. ginny has 12 more than three times the number of baseball cards Leo

marin [14]

Alex has 128 baseball cards, Leo has 66 cards, and ginny has 210.

6 0

3 years ago