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3 years ago

2. Find the probability that an event E will not occur if

Mathematics

1 answer:

Monica [59]3 years ago

3 0

Answer:

Step-by-step explanation:

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Which expressions are equivalent to the one below? Check all that apply.

dalvyx [7]

5^3(5^x) the rule is a^b(a^c)=a^(b+c)

5^(3+x)

8 0

3 years ago

Find the area D (1,3) F (-4,-3) E (4, -3)

tresset_1 [31]

Answer:

Step-by-step explanation:

FE=4+4=8

Altitude=3+3=6

area=1/2×FE×6=1/2×8×6=24 sq. units.

area=1/2×

I 1 3|

|-4 -3|

| 4 -3|

| 1 3|

=1/2[(-3+12)+(12+12)+(12+3)]

=1/2[9+24+15]

=1/2[48]

=24 sq. units

7 0

3 years ago

Please help asap, shehhrhd

Dovator [93]

Answer:

55°

Step-by-step explanation:

y+x=180°

25°+30°+x=180°

25°+30°=y

y=55°

*Exterior Angle Theorem*

<u><em>Sorry for the lack of explanation. It's midnight, so I didn't have much time.</em></u>

7 0

3 years ago

(3^2+4−2)(2^2−3−4) can you guys please solve this for me please.

Degger [83]

Answer:

-33

Step-by-step explanation:

$(3^2+4-2)(2^2-3-4)$

$(9+4-2)(4-3-4)$

$(13-2)(1-4)$

$(11)$ · $(-3)$

(-33)

6 0

3 years ago

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0

3 years ago