Plzzz help need it right nowww.....

Use the hundredths grid to answer the question.

nydimaria [60]

9514 1404 393

Answer:

C. 0.36 +0.27 = 0.63

Step-by-step explanation:

3 columns and 6 squares of a hundredths grid is a representation of ...

0.30 +0.06 = 0.36

2 columns and 7 square is a representation of ...

0.20 +0.07 = 0.27

The total shaded area is the sum of these:

0.36 +0.27 = 0.63

5 0

2 years ago

Three times a number, added to 9 is 33 find the number

kozerog [31]

3 x 8 = 24
24 + 9 = 33

"the number" = 8

6 0

2 years ago

Read 2 more answers

Find the distance if A(2,3) B(4,5)

Ainat [17]

Hii!

=====================================

$\leadsto\mid\boldsymbol{-Answer-}\mid\gets$

Distance =

$\leadsto\mid\boldsymbol{-Explanation-}\mid\gets$

Use the distance formula, which is.

$\sf d=\sqrt{(x2-x1)^2+(y2-y1)^2}$

Plug in the values.

$\sf d=\sqrt{(4-2)^2 +(5-3)^2$

Simplify.

$\sf d=\sqrt{2^2+2^2}$

Simplify.

$\sf d=\sqrt{8}$ .

Decimal Form. 2.828427...

Great job!

--

Hope that this helped! Best wishes.

$\textsl{Reach far. Aim high. Dream big.}\\\boldsymbol{-Greetings!-}$

--

6 0

1 year ago

Read 2 more answers

A box contain 12 balls in which 4 are white 3 are blue and 5 are red.3 balls are drawn at random from the box.find the chance th

Anastasy [175]

Answer:

3/11

Step-by-step explanation:

From the above question, we have the following information

Total number of balls = 12

Number of white balls = 4

Number of blue balls = 3

Number of red balls = 5

We solve this question using combination formula

C(n, r) = nCr = n!/r!(n - r)!

We are told that 3 balls are drawn out at random.

The chance/probability of drawing out 3 balls = 12C3 = 12!/3! × (12 - 3)! = 12!/3! × 9!

= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/(3 × 2 × 1) × (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)

= 220 ways

The chance of selecting 3 balls at random = 220

To find the chance that all the three balls are selected,

= [Chance of selecting (white ball) × Chance of selecting(blue ball) × Chance of selecting(red balls)]/ The chance/probability of drawing out 3 balls

Chance of selecting (white ball)= 4C1

Chance of selecting(blue ball) = 3C1 Chance of selecting(red balls) = 5C1

Hence,

= [4C1 × 3C1 × 5C1]/ 220

= 60/220

= 6/22

= 3/11

The chance that all three are selected is = 3/11

4 0

3 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0

2 years ago