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2 years ago

9

Which function is shown in the graph below?

2 answers:

Alexandra [31]2 years ago

7 0

Answer:

Option C.

Step-by-step explanation:

There are three points which are passing through the logarithmic function shown in the graph.

The points are (0.4, -0.6), (1, 0) and (3, 1).

Now we will plug in the values (3, 1) in all the options to get the correct answer.

a). $y=log_{0.4}x$

We can rewrite the equation as

$(0.4)^{y}=x$ [If $a^{b}=c$ then $b=log_{a}c$ ]

For (3, 1)

$(0.4)^{1}=3$

Which is not true so option A is not the answer.

b). $y=log_{1}x$

We can rewrite the equation as

$(1)^{y}=x$

For (3, 1)

$(1)^{3}=3$

1 = 3

Which is not true.

Therefore, Option B is not the answer.

c). $y=log_{3}x$

We can rewrite the equation as

$(3)^{y}=x$

For (3, 1)

$(3)^{1}=3$

3 = 3

Therefore, option C is the answer.

d). $y=log_{10}x$

We can rewrite the equation as

$(10)^{y}=x$

For (3, 1)

$(10)^{1}=3$

Not true. Therefore, option D is not the answer.

Stolb23 [73]2 years ago

3 0

The annswer is y=log3x

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

Pachacha [2.7K]

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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A sewing club is making a quilt consisting of 25 squares with each side of the square measuring 30 centimeters. If the quilt has

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inysia [295]

$6 \times (4 \times x + y)$

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2 years ago

Question 1: Classify the system of equations and identify the number of solutions. 3 = 4x + y 2y = 6 − 8x Answers: inconsistent;

Mariana [72]

Answer:

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2 years ago

Find two complex numbers that have a sum of i10 a different of -4and a product of -29

lianna [129]

<h2>-2+5i and 2+5i</h2>

Step-by-step explanation:

Let the complex numbers be $a+ib\textrm{ and }c+id$ .

Given, sum is $10i$ , difference is $-4$ and product is $-29$ .

$(a+c)+i(b+d)=10i$ ⇒ $a+c=0,b+d=10$

$(a-c)+i(b-d)=-4$ ⇒ $a-c=-4,b-d=0$

$a=-2,c=2,b=5,d=5$

$(a+ib)(c+id)=(-2+5i)(2+5i)=-4-25=-29$

Hence, all three equations are consistent yielding the complex numbers $-2+5i\textrm{ and }2+5i$ .

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2 years ago