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icang [17]
3 years ago
6

Find the sum of interior angles a polygon has with the given number of sides a.9 b.19

Mathematics
1 answer:
Gnom [1K]3 years ago
4 0

The equation for this is (n-2) x 180

Replace n with the number of sides:

a. 9:  (9-2) x 180 = 7 x 180 = 1,260

b. 19: (19-2) x 180 = 17 x 180 = 3,060

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3 years ago
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The width and length of a rectangle (in feet)are consecutive odd integers. If the length is increased by 5 feet, the area of the
Tresset [83]
Original:
<span>The width and length of a rectangle are consecutive odd integers
</span>so W = x and L = x + 2

<span>If the length is increased by 5 feet, then new L = x + 2 + 5 = x + 7
</span>
A = L x W
60 = (x + 7) x
60 = x^2 + 7x
x^2 + 7x - 60 = 0
(x - 5)(x + 12) = 0
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From here you have x = W = 5 ft and L = x + 2 = 5 + 2 = 7 feet

Area of original = 5 x 7 = 35

answer 
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4 0
3 years ago
Find the geometric mean of the pair of numbers.
tatyana61 [14]
The formula for geometric mean is  nth root  of the product of all terms. n denotes the number of data. Since you have only two points, 99 and 11, n = 2. The product of both terms is 99 × 11 = 1089. The square root of 1089 is 33. Thus, the answer is 33.
6 0
3 years ago
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Assume a researcher recruits 150 African American and Caucasian individuals taking warfarin to determine if there is a differenc
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Answer:

Z=11.0

Step-by-step explanation:

Let Caucasian individuals be the population 1 and African American be the population 2. So, we are given that

n=150

n1=75

mean1=xbar1=6.1.

standard deviation1=S.D1=σ1=1.1.

Variance1=V(x1)=σ1²=1.1²=1.21.

n2=75.

mean2=xbar2=4.3.

standard deviation2=S.D2=σ2=0.9.

Variance2=V(x2)=σ2²=0.9²=0.81.

The z-statistic is

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Z=\frac{6.1-4.3}{\sqrt{\frac{1.21}{75}+ \frac{0.81}{75} } }

Z=\frac{1.8}{\sqrt{0.0161+ 0.0108 } }

Z=\frac{1.8}{\sqrt{0.0269 } }

Z=\frac{1.8}{0.164 } }

Z=10.97

Z=11.0

So, the z value obtained while calculating the test statistic is approximately 11.0.

4 0
2 years ago
Nine is more than twice n.
ella [17]
I hope this helps you


9=n.n

3.3=n.n

n=3
6 0
3 years ago
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