M with the exponent 2 written as an equivalent expression

Fill in the box with the appropriate measure. quick please I have only a few min left

Art [367]

Step-by-step explanation:

first box on the left

r=4

d=8

circumference= 2π*4= 8π

area = π*4*4= 16π

Second box on the left

d=6

r= 3

circumference= 2π*3= 6π

area =π*3*3= 9π

third box on the left

A=36π

A=36πarea= π*r*r

A=36πarea= π*r*rr= 6

A=36πarea= π*r*rr= 6d=12

A=36πarea= π*r*rr= 6d=12circumference= 2π*6= 12 π

the last box

C=18π

C=18πC= 2π*r

C=18πC= 2π*rr= 9

C=18πC= 2π*rr= 9d=18

C=18πC= 2π*rr= 9d=18area= π*9*9= 81π

8 0

2 years ago

The Wall Street Journal Corporate Perceptions Study 2011 surveyed readers and asked how each rated the Quality of Management and

kykrilka [37]

Answer:

The calculated χ² = 17.0281 falls in the critical region χ² ≥ 9.49 so we reject the null hypothesis that the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent

The p-value is 0 .001909. The result is significant at p < 0.05

Part b:

40 > 8.5

35> 7.5

25> 4

Step-by-step explanation:

1) Let the null and alternative hypothesis as

H0: the quality of management and the reputation of the company are independent

against the claim

Ha: the quality of management and the reputation of the company are dependent

2) The significance level alpha is set at 0.05

3) The test statistic under H0 is

χ²= ∑ (o - e)²/ e where O is the observed and e is the expected frequency

which has an approximate chi square distribution with ( 3-1) (3-1)= 4 d.f

4) Computations:

Under H0 ,

Observed Expected E χ²= ∑(O-e)²/e

40 35.00 0.71

25 24.50 0.01

5 10.50 2.88

35 40.00 0.62

35 28.0 1.75

10 12.00 0.33

25 25.00 0.00

10 17.50 3.21

15 7.50 7.50

∑ 17.0281

Column Totals 100 70 30 200 (Grand Total)

5) The critical region is χ² ≥ χ² (0.05)2 = 9.49

6) Conclusion:

The calculated χ² = 17.0281 falls in the critical region χ² ≥ 9.49 so we reject the null hypothesis that the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent.

7) The p-value is 0 .001909. The result is significant at p < 0.05

The p- values tells that the variables are dependent.

Part b:

If we take the excellent row total = 70 and compare it with the excellent column total= 100

If we take the good row total = 70 and compare it with the good column total= 80

If we take the fair row total = 50 and compare it with the fair column total= 30

The two attributes are said to be associated if

Thus we see that ( where (A)(B) are row and columns totals and AB are the cell contents)

AB> (A)(B)/N

40 > 1700/200

40 > 8.5

35> 1500/200

35> 7.5

25> 800/200

25> 4

and so on.

Hence they are positively associated

8 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPP

amid [387]

30

I DONT KNOW BUT THAT'S MY ANSWER HOPE IT HELP

6 0

3 years ago

Read 2 more answers

Listed below are prices in dollars for one night at different hotels in a certain region. Find the range, variance, and standar

Artyom0805 [142]

Answer:

Range = 115$

Standard Deviation = 43.76$

Variance = 1915.142$

Option A) The measures of variation are not very useful because when searching for a room, low prices, location, and good accommodations are more important than the amount of variation in the area.

Step-by-step explanation:

We are given the data for prices in dollars for one night at different hotels in a certain region.

234, 160, 119, 131, 218, 207, 146, 141

Range:

Sorted data: 119, 131, 141, 146, 160, 207, 218, 234

$\text{Range} = 234-119 = 115\$$

Standard Deviation:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1356}{8} = 169.5$

Sum of squares of differences = 4160.25 + 90.25 + 2550.25 + 1482.25 + 2352.25 + 1406.25 + 552.25 + 812.25 = 13406

$\sigma = \sqrt{\dfrac{13406}{7}} = 43.76\$$

Variance =

$\sigma^2 = 1915.142\$$

Measure of variance for someone searching for room:

Option A) The measures of variation are not very useful because when searching for a room, low prices, location, and good accommodations are more important than the amount of variation in the area.

5 0

3 years ago

Day 4: As a landscaper, one of your clients wants a rectangular garden in which the length is five feet

liberstina [14]

Answer:

The Area of Rectangular Garden is 1044 feet²

Step-by-step explanation:

According to question

The perimeter of the garden = 82 ft

Let the length be L ft

The width be W ft

Now as per question

L = 5 + ( 2× W )

∵ Perimeter of Rectangle = 2 × ( Length + Width )

Or , Perimeter of Rectangle = 2 × ( L+ W )

Or, 82 = 2 × ( L+ W )

Or, 82 = 2 × [ 5 + ( 2 ×W ) + W ) ]

Or, 82 = 2 × ( 5 +3W )

Or, 41 = 5 + 3W

Or, 41 - 5 = 3W

So, 3W= 36

∴ W = $\frac{36}{3}$ = 12 feet

I.e Width = 12 feet

And L = 5 + ( 2× W )

Or, Length = 5 + 24 = 29 feet

Now The Area of Rectangle = Length × width

So, The Area of Rectangle = 29 ft × 36 ft

The Area of Rectangle is 1044 feet²

Hence The Area of Rectangular Garden is 1044 feet² Answer

4 0

3 years ago