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2 years ago

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Anna wants to estimate the mean number of siblings for each student in her school. She records the number of siblings for each o

f 75 randomly selected students in the school. What is the statistic?a. the specific number of siblings for each randomly selected student
b. all the students in the school
c. the mean number of siblings for all students in the school
d. the mean number of siblings for the randomly selected students
e. the 75 randomly selected students

1 answer:

Sonbull [250]2 years ago

6 0

Answer:

c and d

Step-by-step explanation:

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A large washer has an outer radius of 10mm and a hole with a diameter of 14mm. What is the area of the top surface of the washer

choli [55]

<h2>Answer:</h2>

The area of the top surface of the washer is: 160.14 square mm.

<h2>Step-by-step explanation:</h2>

The top of the surface is in the shape of a annulus with a outer radius of 10 mm and a inner radius of 7 mm ( since the diameter of the hole is: 14 mm and we know that the radius is half of the diameter)

Now, we know that the area of the annulus region is given by:

$Area=\pi (R^2-r^2)$

where R is the outer radius and r is the inner radius.

Here we have:

$R=10\ mm\\\\and\\\\r=7\ mm$

Hence, we have:

$Area\ of\ top\ surface=\pi (10^2-7^2)\\\\i.e.\\\\Area\ of\ top\ surface=\pi (100-49)\\\\i.e.\\\\Area\ of\ top\ surface=\pi\cdot 51\\\\i.e.\\\\Area\ of\ top\ surface=160.14\ mm^2$

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2 years ago

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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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2 years ago

A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year

uysha [10]

Answer:

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

Step-by-step explanation:

We use two binomial trials to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Probability of at least nine participants finishing the study in a group.

0.2 probability of a students dropping out. So 1 - 0.2 = 0.8 probability of a student finishing the study. This means that $p = 0.8$ .

10 students, so $n = 10$

We have to find:

$P(X \geq 9) = P(X = 9) + P(X = 10)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{10,9}.(0.8)^{9}.(0.2)^{1} = 0.2684$

$P(X = 10) = C_{10,10}.(0.8)^{10}.(0.2)^{0} = 0.1074$

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.2684 + 0.1074 = 0.3758$

0.3758 probability that at least nine participants complete the study in a group.

Calculate the probability that at least nine participants complete the study in one of the two groups, but not in both groups?

0.3758 probability that at least nine participants complete the study in a group. This means that $p = 0.3758$

Two groups, so $n = 2$

We have to find P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,1}.(0.3758)^{1}.(0.6242)^{1} = 0.4691$

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

5 0

2 years ago

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Pedro has created the function f(x)= 4x-3/2 to represent the number of assingments he has completed where x represents the numbe

Law Incorporation [45]

The given function is
f(x) = 4x - 3/2
where
f(x) = number of assignments completed
x = number of weeks required to complete the assignments

We want to find f⁻¹ (30) as an estimate of the number of weeks required to complete 30 assignments.
The procedure is as follows:

1. Set y = f(x)
y = 4x - 3/2

2. Exchange x and y
x = 4y - 3/2

3. Solve for y
4y = x + 3/2
y = (x +3/2)/4

4. Set y equal to f⁻¹ (x)
f⁻¹ (x) = (x + 3/2)/4

5. Find f⁻¹ (30)
f⁻¹ (30) = (30 + 3/2)/4 = 63/8 = 8 (approxmately)

Answer:
Pedro needs about 8 weeks to complete 30 assignments.

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2 years ago

Which expression is equivalent to [4mn/m⁻2n⁻6}⁻2? Assume m and n do not equal zero

leva [86]

The answer would be {(n^5)^2}/{(4*m^3)^2} = { n^{10} }/{16*m^6}

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