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2 years ago

A spinner and 4 cards are shown below:

Mathematics

2 answers:

elena55 [62]2 years ago

7 0

Answer:

1 over 20

Step-by-step explanation:

Given that:

a spinner and 4 cards

A Spinner ⇒ 5 equal sectors;Color: Blue,Purple,Yellow,Red,Green

Arrow point at purple

Four cards ⇒ Color : Red,Yellow,Blue,Pink

Jane spins the spinner & selects card without looking.

To Find - Probability spinner stops at purple & select pink card.

Solve:

Multiply 4 (the number of cards) with 5 (The number of equal sectors)

$\frac{1}{5}\times\frac{1}{4}=\frac{1}{20}$ {1 × 1 = 1} {5×4=20}

~lenvy~

Mama L [17]2 years ago

7 0

Answer:

1 over 20

Step-by-step explanation:

Given the following:

1 spinner and 4 cards

A Spinner which has 5 equal sectors;Color: Green, Blue,Purple,Yellow,Red.

{Arrow point at purple}

Four cards ⇒ which color are Red,Yellow,Blue,Pink

Jane spins the spinner & selects card without looking.

To Find → Probability spinner stops at purple & select pink card.

Since, purple and pink only has one in each of Spinner/card

Thus,

Spinner (5 equal sectors - 1/5}

Card {4 cards - 1/4}

1/4 × 1/5 = 1/20

Hence, probability spinner stops at purple & select pink card is [A] 1/20

[RevyBreeze]

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Answer:

a) $E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

b) $E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$

Step-by-step explanation:

We assume the following distribution given:

Y 0 1 2 3

P(Y) 0.60 0.25 0.10 0.05

Part a

We can find the expected value with this formula:

$E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

Part b

If we want to find the expected value of $100 Y^2$ we need to find the expected value of Y^2 and we have:

$E(Y^2) = \sum_{i=1}^n Y^2_i P(Y_i)$

And replacing we got:

$E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$

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3 years ago

Select two fractions that can be rewritten with a denominator of 24

vagabundo [1.1K]

2/12 and 3/8 Hope it's right YAYYYYY

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3 years ago

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

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lora16 [44]

Answer: Assuming the a2 is a with an exponent of 2 and a3 is a with an exponent of 3, the answer is 2a^6-2a^5

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3 years ago

Which equation represents a line that passes through the two points in the table x 1,5 and y 1,6

aleksandrvk [35]

For this case we must find a linear equation of the form:

$y = mx + b$

Where,

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For the slope we have the following equation:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}}$

Substituting values we have:

$m = \frac {6-5} {1-1}\\m = \frac {1} {0}$

We observe that the slope of the line is not defined.

Therefore, the line is vertical.

Thus, the equation of the line is given by:

$x = 1$

Answer:

The equation of the line is given by:

$x = 1$

8 0

3 years ago

Read 2 more answers