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3 years ago

11

What’s the probability?

2 answers:

tatuchka [14]3 years ago

7 0

Answer:

sorry I don't know

Step-by-step explanation:

OK

OK jsjzjxjxndzkxk

s

PLZ MARK ME AD BRAINLIST

jek_recluse [69]3 years ago

4 0

<h3>Answer:</h3>

<u>We know that</u>,

<u>P(A) = P(B|A) - P(A ∩ B)</u>

3/10 - 21/100
30 - 21/100
7/100

<u>Hence, P(A) is 7/100</u>.

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Convert improper fraction to a mixed number<br> 9/2

harkovskaia [24]

2 goes into 9 4 times. So 4 is your large number and you have 1 remaining, put that over the denominator and your answer is 4 1/2

8 0

3 years ago

A person takes up 24 inches of space on a stadium bleacher. If one row of bleachers is 150 feet long and there are 20 rows of bl

Feliz [49]

Answer:

1,500 people

Step-by-step explanation:

Remember that

$1\ ft= 12\ in$

Convert ft to in

$150\ ft=150*12=1,800\ in$

<em>Divide the length of a row by the space occupied by one person</em>

$1,800/24=75\ persons$

<em>Multiply by the number of rows to determine the total people that can fit into the stands</em>

$75(20)=1,500\ people$

3 0

3 years ago

Evaluate the function f(x) = -7x + 2 at f(-18).

enyata [817]

Answer:

128

Step-by-step explanation:

-7x+2 = -18(-7)+2=126+2=128

7 0

2 years ago

Please Help Me I Need Finish This Tonight

vladimir2022 [97]

1) a- rectangle
b- same place you started, facing west
2) a- right triangle
b- parallelogram
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5 0

3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago