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Anastaziya [24]

3 years ago

13

4. (10 pts) A study is conducted to estimate the difference in the mean occupational exposure to radioactivity in computer scien

ce students in the years 2000 and 2008. These data based on independent samples of students for the 2 years are obtained:

1 answer:

IceJOKER [234]3 years ago

5 0

Answer:

ddd

Step-by-step explanation:

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The area of a triangle is 10 square meters. The height is one meter longer than the base. What are the lengths of the base and h

Kamila [148]

Answer:

b=4, h=5

Step-by-step explanation:b(b+1) divided by 2 is 10, so b(b+1)=20 so b^2 + b =20 so b^2 +b -20=0 which equals (b+5)(b-4), giving us solutions -5 and 4. However, -5 is invalid bc it is negative, so base must be 4 making height 5

4 0

3 years ago

Can you help don’t understand

densk [106]

7^4 * 4^7

hope this helps

8 0

3 years ago

What is the value of x?<br> A. 13 <br> B. 17<br> C. 119<br> D. 169

Marianna [84]

Answer:

13

Step-by-step explanation:

x² = 5²+12²

x² = 25+144

x² = 169

x² = 13²

x = 13

3 0

3 years ago

2/3 of 27 is equal to what fraction?

horrorfan [7]

<u>2</u> <u>27</u>
3 times 1

<u>2 times 27 = 54</u>
3 times 1 = 3

54 divided by 3 = 18

<em>TIP: The fraction bar is also used as a division sign.</em>

7 0

4 years ago

Read 2 more answers

Write an equation for an ellipse centered at the origin, which has foci at (\pm5,0)(±5,0)(, plus minus, 5, comma, 0, )and vertic

Virty [35]

Answer:

The equation of ellipse is

$\frac{x^2}{41}+\frac{y^2}{16}=1$

Step-by-step explanation:

The equation of an ellipse is

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

where(h,k) is the center and c is distance from the center to the foci is given by $a^2-b^2=c^2$ . a is the distance from the center to the vertices and b is the distance from the center to the co-vertices.

The center of the ellipse is the mid-point of the vertices.

The mid point of the vertices $(\pm\sqrt{41},0)$ is

= $(\frac{\sqrt{41}+(-\sqr{41})}2,\frac{0+0}2)$

=(0,0)

a is the distance between the center and the vertices.

So, $a=\sqrt{(0-\sqrt{41})^2+(0-0)^2}$

$=\sqrt{41}$

c is the distance between the center and the foci.

So, $c=\sqrt{(0-5)^2+(0-0)^2$

=5

$a^2-b^2=c^2$

$\Rightarrow (\sqrt41)^2-b^2=5^2$

$\Rightarrow b^2=(\sqrt41)^2- 5^2$

$\Rightarrow b^2=41-25$

$\Rightarrow b^2=16$

The equation of ellipse is

$\frac{(x-0)^2}{(\sqrt{41})^2}+\frac{(y-0)^2}{16}=1$

$\Rightarrow \frac{x^2}{41}+\frac{y^2}{16}=1$

7 0

4 years ago