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3 years ago

11

Jessica is at a charity fundraiser and has a chance of receiving a gift. The odds in favor of receiving a gift are 5/12. Find th

e probability of Jessica receiving a gift.

1 answer:

Goryan [66]3 years ago

3 0

Answer:

5/17

Step-by-step explanation:

This is a question to calculate probability from odds. The formula is given as:

A formula for calculating probability from odds is P = Odds / (Odds + 1)

From the question , we are told that the odds of receiving a gift is

= 5:12

The probability of Jessica receiving a gift =

Probability = Odds / (Odds + 1)

P = 5/12 / ( 5/12 + 1)

P = (5/12)/ (17/12)

P = 5/12 × 12/17

= 5/17

Therefore, the probability of Jessica. receiving a gift is 5/17.

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Teddy has two piggy banks. The difference in the amount of money between the two banks is no more than $10. One piggy bank has $

just olya [345]

Answer:

2.69≤x≤17.31

Step-by-step explanation:

Lety the amount in the other Piggy bank be x.

The difference in the amount of money between the two banks is no more than $10 is expressed as;

|x-7.31| ≤ 10

The value in the modulus sign can be positive or negative

If positive;

x-7.31 ≤ 10

x ≤ 10+7.31

x≤17.31

If negative

-(x-7.31) ≤ 10

-x+7.31 ≤ 10

-x ≤ 10-7.31

-x ≤ 2.69

Multiply both sides by -1

x ≥ -2.69

The compound inequality representing the amount of money in the other bank is -2.69≤x≤17.31

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2 years ago

The lifespan of a guinea pig is 6 years less than that of a giraffe. The lifespan of a tiger is 4 times that of the guinea pig.

Ronch [10]

Answer:

Giraffe = 10

Guinea Pig = 4

Tiger = 16

Step-by-step explanation:

G GP T = 30

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X X-6 (X-6)*4

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9 3 12 = 24

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10 4 16 = 30 ✅

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2 years ago

Determine the type and number of solutions of 6x^2-7x-4=0.

inessss [21]

2 solutions (X^2)
The Answer is D. Two Real Solutions

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3 years ago

The rectangle below has an area of 6n^4+20n^3+14n^26n 4 +20n 3 +14n 2 6, n, start superscript, 4, end superscript, plus, 20, n,

Sindrei [870]

Given:

Area of rectangle = $6n^4+20n^3+14n^2$

Width of the rectangle is equal to the greatest common monomial factor of $6n^4, 20n^3,14n^2$ .

To find:

Length and width of the rectangle.

Solution:

Width of the rectangle is equal to the greatest common monomial factor of $6n^4, 20n^3,14n^2$ is

$6n^4=2\times 3\times n\times n\times n\times n$

$20n^3=2\times 2\times 5\times n\times n\times n$

$14n^2=2\times 7\times n\times n$

Now,

$GCF(6n^4, 20n^3,14n^2)=2\times n\times n=2n^2$

So, width of the rectangle is $2n^2$ .

Area of rectangle is

$Area=6n^4+20n^3+14n^2$

Taking out GCF, we get

$Area=2n^2(3n^2+10n+7)$

We know that, area of a rectangle is the product of its length and width.

Since, width of the rectangle is $2n^2$ , therefore length of the rectangle is $(3n^2+10n+7)$ .

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3 years ago

Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−

aleksandrvk [35]

The tangent to $C$ through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces $S$ and $T$ at that point.

Let $f(x,y,z)=x^2+2y^3+3z^4$ . Then $S$ is the level curve $f(x,y,z)=6$ . Recall that the gradient vector is perpendicular to level curves; we have

$\nabla f(x,y,z)=(2x,6y,12z^2)$

so that the gradient of $f$ at (1, 1, 1) is

$\nabla f(1,1,1)=(2,6,12)$

For the surface $T$ , we have

$\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0$

so that $\vec r(1,0)=(1,1,1)$ . We can obtain a vector normal to $T$ by taking the cross product of the partial derivatives of $\vec r(u,v)$ , and evaluating that product for $u=1,v=0$ :

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)$

$\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)$

Now take the cross product of the two normal vectors to $S$ and $T$ :

$(2,6,12)\times(1,-1,2)=(24,8,-8)$

The direction of vector (24, 8, -8) is the direction of the tangent line to $C$ at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by $t\in\Bbb R$ . Then adding (1, 1, 1) shifts this line to the point of tangency on $C$ . So the tangent line has equation

$\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)$

7 0

3 years ago