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dybincka [34]
3 years ago
5

find the area of the following shape. A(-5,-8) B(1,-8) C(3,-5) D(1,0) E(-5,-3) F(-3,-6). just show work

Mathematics
1 answer:
Juliette [100K]3 years ago
6 0
Check the picture below


so hmmm, all you really have is 4 triangles and 1 rectangle, notice the bases and altitude of each triangle

so, get the area of each, sum them up, and that's the area of the shape

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HURRY ONLY HAVE 1 MIN!!!!!!!!!!!!!!!!
BaLLatris [955]

Answer:

100 m3

Step-by-step explanation:

2*5*10 = 10*10 *100

6 0
2 years ago
A. 18 square root of 2 B.18/ square root of 3 C.18 square root of 2 / square root of 3 D. 18 times square root 3 / square root o
kirza4 [7]

Step-by-step explanation:

A. 18√2

= 25.4558441227

B. 18/√3

= 6√3

= 10.39230485

C. 18√2 /√3

= 6√6

= 14.69693846

D. 18 · √3 / √2

= 9√6

= 22.04540769

6 0
3 years ago
A random sample of 85 sixth-graders in a large city take a course designed to improve scores on a reading comprehension test. Ba
BartSMP [9]

Answer:

12.6 < \mu < 14.8

For this case we can find the margin of error like this:

ME= \frac{14.8-12.6}{2}= 1.1

Since we need to divide the width of the interval by 2.

And now with the margin of error we can find the sample mean with any of the two following ways:

12.6 +1.1=13.7

14.8-1.1=13.7

So for this case the correct answer would be:

A. Sample Mean = 13.7; Margin of Error = 1.1

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=85-1=84

We know that the confidence interval is given by:

12.6 < \mu < 14.8

For this case we can find the margin of error like this:

ME= \frac{14.8-12.6}{2}= 1.1

Since we need to divide the width of the interval by 2.

And now with the margin of error we can find the sample mean with any of the two following ways:

12.6 +1.1=13.7

14.8-1.1=13.7

So for this case the correct answer would be:

A. Sample Mean = 13.7; Margin of Error = 1.1

3 0
3 years ago
A total of $37185 is invested at an annual interest rate of 2% compounded quarterly. Find the balance after 11 years.
vazorg [7]
A=37,185×(1+0.02÷4)^(4×11)
A=46,309.97
8 0
3 years ago
Please I need someone to help my assignment​
Nadusha1986 [10]
(x-4)² + (x-2)² = x²

(x² - 8x + 16 ) + (x² - 4x + 4) = x

2x² -12x + 20 = x

2x²-12x + 20 - x = 0

2x² - 13x + 20 = 0

(2x-5) (x-4) = 0

2x-5 = 0

2x = 5

x = ⁵/₂


x-4 = 0

x = 4



so x₁ = ⁵/₂ , and x₂ = 4
4 0
2 years ago
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