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xeze [42]
3 years ago
9

Square root of -20 someone please help

Mathematics
1 answer:
postnew [5]3 years ago
4 0

Answer:

I think its either 5 or 4 one of the 2

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The base of a triangle is fixed at 2.218 millimeters. Determine the number of
harina [27]

When the base of a triangle is fixed at 2.218 millimeters, the area will be 0.1098 mm².

<h3>What will the area be?</h3>

It should be noted that the area of a triangle is calculated as:

Area of triangle = (1/2)×base×height

= (1/2) × 2.218 mm × 0.099 mm

= 1.109 mm × 0.099 mm

= 0.109791 mm²

≈ 0.1098 mm²

Therefore, the <em>area</em> is 0.1098 mm².

Learn more about triangles on;

brainly.com/question/2644832

#SPJ1

The base of a triangle is fixed at 2.218 millimeters. Determine the number of

significant figures of the area of the triangle with each given height of 0.099mm.

4 0
1 year ago
21+[36devided(12-6)+2-5]multiply3
Alona [7]
12-6=6
6+2=8-5=3
21+3x3=72
4 0
3 years ago
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PLEASE HELP find the value of x
Elenna [48]

I think it could be 40° or 130°. Im not 100% sure however because i dont know how much all the numbers add up to. It could either be 180° or 270°

7 0
3 years ago
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I need help. the Question in in Photo
7nadin3 [17]

Answer:

I had same one so the answer is C

Step-by-step explanation:

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6 0
3 years ago
Wind farm configuration is a significant issue as the upwind turbines generate maximal energy while creating wakes for downwind
lord [1]

Answer:

a) [24.114,29.2858]

b) Since 30 is not in the 95% confidence interval, there is a 95% probability that 30 is not the true mean WEP

Step-by-step explanation:

a)

The 95% confidence interval is given by the interval

\bf [ \bar x-z^*\frac{s}{\sqrt n}, \bar x+t^*\frac{s}{\sqrt n}]

where

\bf \bar x= 26.7 is the sample mean  

s = 17.7 is the sample standard deviation  

n = 180 is the sample size

Since the sample size is big enough, we can use the Normal N(0,1) to compute \bf z^* and it would be 1.96(*) (a value such that the area under the Normal curve outside the interval [-z, z] is 5% (0.05))

and our 95% confidence interval is

\bf [26.7-1.96*\frac{17.7}{\sqrt{180}}, 26.7+1.96*\frac{17.7}{\sqrt{180}}]=\boxed{[24.114,29.2858]}

(*)

This value can be computed in Excel with

<em>NORMINV(1-0.025,0,1)</em>

and in OpenOffice Calc with

<em>NORMINV(1-0.025;0;1)</em>

b)

Since 30 is not in the 95% confidence interval, there is a 95% probability that 30 is not the true mean WEP

7 0
3 years ago
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