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Yuri [45]
3 years ago
14

Help mathh PLEAEE HELP​

Mathematics
1 answer:
Bess [88]3 years ago
8 0

this is as far as i could get. sorry

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A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will h
podryga [215]

Answer:

a) y </= 8 is not sufficient

b) y >/= 4 is sufficient (y is not less than 3.5)

Step-by-step explanation:

Number of Red marbles = 8

Number of white marbles = y

Total number of marbles in the jar = 8+ y

Let Pr(R) be the probability of picking red marbles

Let Pr(W) be the probability picking white marbles

Pr(R) = 8/ (8+y)

Pr(W) = 7/(7+y)

Pr(RR) = Pr(R1) * Pr( R2)

= 8/(8+y) * 7/(7+y)

Pr(RW) = Pr(R1) * Pr(W2) + Pr(W1) * Pr(R2)

= 2[Pr(R1) * Pr(W2)

= 2[8/(8+y) * y/(7+y)]

The probability of having 2 red is greater than one marble of each color.

Pr(RR) > Pr( RW)

8/(8+y) * 7/(7+y) > 2[8/(8+y) * y/(7+y)]

7/(7+y) > 2(y/(7+y)

7/y > 2

7/2 > y

3.5 > y

y < 3.5

Therefore;

a) y </= 8 is not sufficient

b) y >/= 4 is sufficient (y is not less than 3.5)

6 0
3 years ago
What is the answer to 11y+13=12x
kherson [118]

Answer:

36

Step-by-step explanation:

7 0
3 years ago
Alexis and Jessica are shopping. Alexis buys 4 pairs of pants and 3 necklaces and pays $216. Jessica buys 7 pairs of pants and 2
kkurt [141]

Step-by-step explanation:

let us keep the price of pants as <em>x</em>

and price of necklace as <em>y</em>

Simultaneous equations comes as follows:

<em>4x</em><em> </em><em>+</em><em> </em><em>3y</em><em> </em><em>=</em><em> </em><em>2</em><em>1</em><em>6</em><em> </em>for Alexis

<em>7x</em><em> </em><em>+</em><em> </em><em>2y</em><em> </em><em>=</em><em> </em><em>3</em><em>0</em><em>0</em><em> </em><em>for</em><em> </em><em>Jess</em>

<em>we'll</em><em> </em><em>make</em><em> </em><em>either</em><em> </em><em>x</em><em> </em><em>or</em><em> </em><em>y</em><em> </em><em>equal</em><em> </em>

<em>here</em><em> </em><em>let's</em><em> </em><em>make</em><em> </em><em>y</em><em> </em><em>equal</em><em> </em>

<em>2</em><em> </em><em>(</em><em> </em><em>4x</em><em> </em><em>+</em><em> </em><em>3y</em><em> </em><em>=</em><em> </em><em>2</em><em>1</em><em>6</em><em>)</em>

<em>3</em><em> </em><em>(</em><em> </em><em>7x</em><em> </em><em>+</em><em> </em><em>2y</em><em> </em><em>=</em><em> </em><em>3</em><em>0</em><em>0</em><em>)</em>

<em>8x</em><em> </em><em>+</em><em> </em><em>6y</em><em> </em><em>=</em><em> </em><em>4</em><em>3</em><em>2</em>

<em>21x</em><em> </em><em>+</em><em> </em><em>6y</em><em> </em><em>=</em><em> </em><em>9</em><em>0</em><em>0</em>

<em>21x</em><em> </em><em>-</em><em> </em><em>8x</em><em> </em><em>=</em><em> </em><em>9</em><em>0</em><em>0</em><em> </em><em>-</em><em> </em><em>4</em><em>3</em><em>2</em>

<em>13x</em><em> </em><em>=</em><em> </em><em>4</em><em>6</em><em>8</em>

<em>x</em><em> </em><em>=</em><em> </em><em>$</em><em>3</em><em>6</em>

<em>so</em><em> </em><em>a</em><em> </em><em>pant</em><em> </em><em>costs</em><em> </em><em>$</em><em>3</em><em>6</em>

<em>And</em><em> </em>

<em>3y</em><em> </em><em>=</em><em> </em><em>7</em><em>2</em>

<em>y</em><em> </em><em>=</em><em> </em><em>$</em><em> </em><em>2</em><em>4</em>

<em>so</em><em> </em><em>a</em><em> </em><em>necklace</em><em> </em><em>costs</em><em> </em><em>$</em><em>3</em><em>6</em>

7 0
3 years ago
The sum of two numbers is -5. The difference is<br> 13. What is the smaller number?
Alex73 [517]

Answer:

The smaller number is -9.

Step-by-step explanation:

Let <em>x</em> and <em>y</em> be the two unknown numbers.

Their sum is -5. Hence:

x+y=-5

Their difference is 13. Hence:

x-y=13

This yields a system of equations. We can solve this using elimination. Adding the two equations together:

(x+y)+(x-y)=(-5)+(13)

Simplify:

2x=8

Divide both sides by two. Thus:

x=4

Using either equation (I'll use the first):

(4)+y=-5\Rightarrow y=-9

So, our two numbers are 4 and -9.

The smaller number is -9.

6 0
3 years ago
Lim x tends to a now find x^1/3-a^1/3/x^1/2-a^1/2
blagie [28]

Answer: \frac{a}{3}-\frac{2}{3}-\frac{a}{2}

Step-by-step explanation:

\lim _{x\to \:a}\left(\frac{x^1}{3}-\frac{\frac{a^1}{3}}{\frac{x^1}{2}}-\frac{a^1}{2}\right)

=\frac{a^1}{3}-\frac{\frac{a^1}{3}}{\frac{a^1}{2}}-\frac{a^1}{2}

=\frac{a}{3}-\frac{2}{3}-\frac{a}{2}

4 0
3 years ago
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