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2 years ago

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What is the simplified form of StartRoot StartFraction 72 x Superscript 16 Baseline Over 50 x Superscript 36 Baseline EndFractio

n EndRoot? Assume x ≠ 0.
StartFraction 6 Over 5 x Superscript 10 Baseline EndFraction
StartFraction 6 Over 5 x squared EndFraction
Six-fifths x Superscript 10
Six-fifths x squared
Mark this and return

1 answer:

aalyn [17]2 years ago

9 0

Answer:

StartFraction 6 Over 5 x Superscript 10 Baseline EndFraction

Step-by-step explanation:

Apparently you want to simplify ...

$\sqrt{\dfrac{72x^{16}}{50x^{36}}}$

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

1/a^b = a^-b

(a^b)^c = a^(bc)

__

So the expression simplifies as ...

$\sqrt{\dfrac{72x^{16}}{50x^{36}}}=\sqrt{\dfrac{36}{25x^{36-16}}}=\sqrt{\dfrac{36}{25x^{20}}}\\\\\sqrt{\left(\dfrac{6}{5x^{10}}\right)^2}=\boxed{\dfrac{6}{5x^{10}}}$

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Which type of isometry is the equivalent of two reflections across intersecting lines?

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B) rotation.

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Penny charges $5 per hour for babysitting, plus a flat fee of $12 for her travel expenses. Charlene charges $3 per hour for baby

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Answer: 5x+12=3x+18

Step-by-step explanation: let x = hours worked

Penny earns $5/hour + $12 fee so 5x+12

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2 years ago

A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L

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Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

$P(X=x)=(nCx)p^{x}(1-p)^{n-x}$

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

$nCx=\frac{n!}{x!(n-x)!}$

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

$P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}$

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

$P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}$

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

$P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}$

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

$P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)$

$P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}$

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

$Y1 + Y2 \leq 1$

$Y1 + Y2\leq 1$ when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate $P(Y1+Y2\leq 1)$ we must sume all the probabilities that satisfy the equation :

$P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)$

$P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448$

$P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224$

$P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192$

$P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864$

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Answer:

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1 : 50

We need the first number to be 18 so multiply by 18

1:18 : 50*18

18 cm : 900 cm

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