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3 years ago

Given h(2) = -3x – 4, find h(0).

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

4 0

Answer:

h(0) = -4

Step-by-step explanation:

Step 1: Define

h(x) = -3x - 4

h(0) is x = 0

Step 2: Substitute and Evaluate

h(0) = -3(0) - 4

h(0) = 0 - 4

h(0) = -4

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The relationship between the countries of the Allied Powers?

Brrunno [24]

Allied powers, also called Allies, those countries allied in opposition to the Central Powers (Germany, Austria-Hungary, and Turkey) in World War I or to the Axis powers (Germany, Italy, and Japan) in World War II. did you know? Romania switched sides and became one of the Allied Powers during WWII in 1944.

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3 years ago

.25% written as a fraction and decimal

Genrish500 [490]

Answer:

1/4 in fraction form, and 0.25 in decimal form.

Step-by-step explanation:

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3 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

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 :

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

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

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

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

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

7 0

3 years ago

Find the fourth term of an arithmetic sequence if the first term is 12 and the common difference is -3. A (subscript 4)=???

storchak [24]

Answer:

The fourth term of the sequence is 3

Step-by-step explanation:

For an nth term in an arithmetic sequence

A(n) = a + (n - 1)d

Where n is the number of terms

a is the first term

d is the common difference

From the question

a = 12

d = - 3

So the fourth term of the sequence is

A(4) = 12 + (4 - 1)-3

= 12 + (3)(-3)

= 12 - 9

The final answer is

Hope this helps you.

7 0

4 years ago

Which box and whisker plot matches the data 20, 32, 19, 12, 28, 34, 29, 36, 20, 15, 30, 17

jasenka [17]

Answer:

See attached images.

Step-by-step explanation:

The minimum of the data set is 12.

Maximum is 36.

Median is 24.

First quartile Q1 = 18

Third quartile Q3 = 31

5 0

3 years ago