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umka21 [38]
3 years ago
14

Prior to hosting an international soccer match, the local soccer club needs to replace the artificial turf on their field with g

rass turf. The grass turf will cost $ 6.75 per square meter. If the field is 0.101 km by 0.065 km, how much will it cost the club to add the grass turf to their field
Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
3 0

Answer:

$44,313.75

Step-by-step explanation:

First, we need to turn the km into meters. There are 1000 meters in a km meaning we need to multiply each of these values by 1000.

0.101 km * 1000= 101 meters

0.065 km * 1000 =  65 meters

Now we need to multiply these values together to find the area of the field

101m * 65m = 6,565m^{2}

Finally, we need to multiply this area by the cost per square meter

6,565m^{2} / 6.75 = $44,313.75

The total repair will cost $44,313.75

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If the legs of a right triangle are both 8 cm, find the area of the triangle. A) 8 cm2 B) 16 cm2 C) 32 cm2 D) 64 cm2
Alex787 [66]

Answer:

32cm

Step-by-step explanation:

Area of a triangle= 1/2 X b X h

b= 8

h=8

1/2 X 8 X 8

= 4 X 8

= 32cm

5 0
3 years ago
Read 2 more answers
Please help. Fill in the blanks with my rule
Aleksandr [31]

Answer:

2 - 9

3 - 16

4 - 25.

Step-by-step explanation:

I think you want this:

2 - (2+1)(2+1)

3 - (3 + 1)(3+1)

4 - (4+1)(4 + 1)

You substitute 2, 3 and 4 into  x - (x+1)(x+1) and work them out.

5 0
2 years ago
public accountant financial planner 50.2 48.0 59.8 49.2 57.3 53.1 59.2 55.9 53.2 51.9 56.0 53.6 49.9 49.7 58.5 53.9 56.0 52.8 51
ddd [48]

Using the formula of test statistic, the value of the test statistic is 2.6961.

The mean of the public accountant is

Mean x(p) =50.2+59.8+57.3+59.2+53.2+56.0+49.9+58.5+56.0+51.9/10

x(p) = 55.2

Now the standard deviation of public accountant is

SD(p) = √{∑(x-x(p))^2/n-1}

SD(p) = √(50.2-55.2)^2+(59.8-55.2)^2+..................+(51.9-55.2)^2/n-1

After solving;

SD(p) = 3.34

The mean of the financial planner is

Mean x(F) =48.0+49.2+53.1+55.9+51.9+53.6+49.7+53.9+52.8+48.9/10

x(F) = 51.6

Now the standard deviation of financial planner is

SD(F) = √{∑(x-x(p))^2/n-1}

SD(F) = √(48.0-51.6)^2+(49.2-51.6)^2+..................+(48.9-51.6)^2/n-1

After solving;

SD(F) = 2.57

Test Statistic (t) = \frac{x(P)-x(f)}{\sqrt{\left(\frac{(n(p)-1)s(p)^2+(n(f)-1)s(f)^2}{n(p)+n(f)-2}\left(\frac{1}{n(p)}+\frac{1}{n(f)}\right)\right)}}

t = \frac{55.2-51.61}{\sqrt{\left(\frac{(10-1)(3.34))^2+(10-1)(2.57)^2}{10+10-2}\left(\frac{1}{10}+\frac{1}{10}\right)\right)}}

After solving

t = 2.6961

Hence, the value of the test statistic is 2.6961.

To learn more about test statistic link is here

brainly.com/question/14128303

#SPJ4

The right question is

public accountant  50.2  59.8  57.3  59.2  53.2  56.0  49.9  58.5  56.0  51.9

financial planner   48.0  49.2  53.1   55.9  51.9   53.6  49.7  53.9 52.8  48.9

Use a 0.05 level of significance and test the hypothesis that there is no difference between the starting annual salaries of public accountants and financial planner

Find the value of the test statistic.

4 0
1 year ago
What are the x-intercepts of the graph of the function f(x)= x^2+4x-12
levacccp [35]
(2,0)
(-6,0)
I just inserted the equation into a graphing calculator. Kinda lazy tonight :/
7 0
3 years ago
2 1/8 - 1 1/4 can someon e help????
olasank [31]
The answer would be: -0.125
7 0
2 years ago
Read 2 more answers
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