g(x)=\dfrac{x+6}{2}g(x)=
2
x+6
​
g, left parenthesis, x, right parenthesis, equals, start fraction, x, plus, 6, divided by, 2, end fraction
What is the domain of g?g

Answer:

[tex]d = \dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]

Step-by-step explanation:

Given the equation of velocity w.r.to time 't':

[tex]v=9t-\dfrac{2}{9}+t^2 ...... (1)[/tex]

Formula for Displacement:

[tex]Displacement = \text{velocity} \times \text{time}[/tex]

So, if we find integral of velocity w.r.to time, we will get displacement.

[tex]\Rightarrow \text{Displacement}=\int {v} \, dt[/tex]

[tex]\Rightarrow \int {v} \, dt = \int ({9t-\dfrac{2}{9}+t^2}) \, dt \\\Rightarrow \int{9t} \, dt - \int{\dfrac{2}{9}} \, dt + \int{t^2} \, dt\\\Rightarrow s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3} + C ....... (1)[/tex]

Here, C is constant (because it is indefinite integral)

Formula for integration used:

[tex]1.\ \int({A+B}) \, dx = \int {A} \, dx + \int{B} \, dx \\2.\ \int({A-B}) \, dx = \int {A} \, dx - \int{B} \, dx \\3.\ \int{x^{n} } \, dx = \dfrac{x^{n+1}}{n+1}\\4.\ \int{C } \, dx = Cx\ \{\text{C is a constant}\}[/tex]

Now, it is given that s = 0, when t = 0.

Putting the values in equation (1):

[tex]0=\dfrac{9\times 0^{2} }{2} - \dfrac{2}{9}\times 0 + \dfrac{0^3}{3} + C\\\Rightarrow C = 0[/tex]

So, the equation for displacement becomes:

[tex]s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]

Answer:

x^2+2x+3 I sent the answer

Answer:

X=1 x=-9

Step-by-step explanation:

(x^2)+8x=9

(x^2)+8x-9=0

(x-1)(x+9)=0

x=1 x=-9

Answer: A

Step-by-step explanation: 5y=x+5 simplified to slope-intercept form is y = 1/5x + 1

The equation 5y = x + 5 is a linear equaton, and the graph A represents the equation 5y = x + 5

The equation is given as:

5y = x + 5

Make x = 0.

So, we have:

5y = 0 + 5

This gives

5y = 5

So, we have:

y = 1

The point is represented as (0,1)

Make y = 0.

So, we have:

5 * 0 = x + 5

This gives

0 = x + 5

So, we have:

x = -5

The point is represented as (-5,0)

This means that the graph of 5y = x + 5 passes through the points (0,1) and (-5,0)

From the list of options, only graph (A) passes through the points (0,1) and (-5,0)

Hence, the graph A represents the equation 5y = x + 5

Read more about linear equations at:

https://brainly.com/question/14323743

#SPJ2

Answer:

(a)20

(b)Elastic

(c)8

(d) Elastic

Step-by-step explanation:

Elasticity of demand(E) indicates the impact of a price change on a product's sales.

The general formula for an exponential demand curve is given as:

[tex]y=ae^{-bp}[/tex]

Given the demand curve formula

[tex]q=f\left(p\right)=200e^{-0.4p}[/tex]

The formula for Elasticity of demand, E

[tex]E = -\dfrac{p}{q}\dfrac{\text{d}q}{\text{d}p}[/tex]

(a)When Price,  p = $50

p=50

[tex]q=200e^{-0.4*50}=200e^{-20}[/tex]

[tex]\dfrac{\text{d}q}{\text{d}p}=-0.4*200e^{-0.4p}=-80e^{-0.4p}[/tex]

Therefore:

[tex]E = -\dfrac{50}{200e^{-20}}*-80e^{-0.4*50}\\=\dfrac{1}{4e^{-20}}*80e^{-20}\\\\E=20[/tex]

(b)At p = $50, Since elasticity is greater than 1, the demand is elastic.

An elasticity value of 20 means that a 1% increase in price causes a 20% decrease in demand.

(c)At p=$20

p=20

[tex]q=200e^{-0.4*20}=200e^{-8}[/tex]

[tex]\dfrac{\text{d}q}{\text{d}p}=-0.4*200e^{-0.4p}=-80e^{-0.4p}[/tex]

Therefore:

[tex]E = -\dfrac{20}{200e^{-8}}*-80e^{-0.4*20}\\=\dfrac{1}{10e^{-20}}*80e^{-20}\\\\E=8[/tex]

(d)At p = $20, the demand is elastic.

An elasticity value of 8 means that a 1% increase in price causes a 8% decrease in demand.

Answer:

88 mins

Step-by-step explanation:

Divide 40 by 5 to get 8. That is your mile per minute ratio. Then multiply 11 by 8 to get the time of 11 miles!

Answer:

There were 89 retail stores in 1997.

Step-by-step explanation:

You know:

y =  11*x + 12

where y is the total number of stores and x is the number of years after 1990.

Being this a linear function (Linear functions are those functions that have the form y = mx + b as it happens in this case), to know the total number of stores and, then you must know the value of x, that is, the number of years after 1990, and replace that value in the given function.

To find out how many retail stores there were in 1997, then since 1990 7 years have passed, calculated by subtracting: 1997 - 1990 = 7

So, 7 is the number of years after 1990, then x=7

Replacing in the function y=11*x + 12 you get:

y=11*7+12

Solving:

y=77+12

y=89

There were 89 retail stores in 1997.

Answer:

18

Step-by-step explanation:

pls mark brainliest

idk if this is a joke

Answer:

Step-by-step explanation:

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Answer:

explanation is below happy to help ya!

Step-by-step explanation:

Answer:

0.126 = 12.6% probability of selling less than 5 properties in one week.

Step-by-step explanation:

For each property, there are only two possible outcomes. The chance of selling any one property is independent of selling another property. So we use the binomial probability distribution 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.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

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

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A real estate agent has 17 properties that she shows.

This means that [tex]n = 17[/tex]

She feels that there is a 40% chance of selling any one property during a week.

This means that [tex]p = 0.4[/tex]

Compute the probability of selling less than 5 properties in one week.

[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{17,0}.(0.4)^{0}.(0.6)^{17} = 0.0002[/tex]

[tex]P(X = 1) = C_{17,1}.(0.4)^{1}.(0.6)^{16} = 0.0019[/tex]

[tex]P(X = 2) = C_{17,2}.(0.4)^{2}.(0.6)^{15} = 0.0102[/tex]

[tex]P(X = 3) = C_{17,3}.(0.4)^{3}.(0.6)^{14} = 0.0341[/tex]

[tex]P(X = 4) = C_{17,4}.(0.4)^{4}.(0.6)^{13} = 0.0796[/tex]

[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0019 + 0.0102 + 0.0341 + 0.0796 = 0.126[/tex]

0.126 = 12.6% probability of selling less than 5 properties in one week.

Answer:

The x-intercepts, or zeros, which are the values of x(or a in this problem) for which the function is 0, are x = a = -2 and x = a = -3.

Step-by-step explanation:

Suppose we have a function y = f(x). The zeros, which are the values of x for which y = 0, are also called the x-intercepts of the function.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

[tex]ax^{2} + bx + c, a\neq0[/tex].

This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:

[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]

[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]

[tex]\bigtriangleup = b^{2} - 4ac[/tex]

In this question:

I will write the function as a function of x, just exchanging a for x.

[tex]f(x) = x^{2} + 5x + 6[/tex]

[tex]\bigtriangleup = 5^{2} - 4*1*6 = 1[/tex]

[tex]x_{1} = \frac{-5 + \sqrt{1}}{2*1} = -2[/tex]

[tex]x_{2} = \frac{-5 - \sqrt{1}}{2*1} = -3[/tex]

The x-intercepts, or zeros, which are the values of x(or a in this problem) for which the function is 0, are x = a = -2 and x = a = -3.

Answer:

5c

Step-by-step explanation:

5c is the only value you can take out of both factors

Answer:

The area is 120 cm²

the perimeter is 52 cm

Step-by-step explanation:

The rhombus area is given by:

A = D * d / 2

that is, the larger diagonal D by the smaller diagonal d, between two, we know that D = 24 cm and d = 10 cm, replacing:

A = 24 * 10/2

A = 120

The area is 120 cm²

To calculate the perimeter use the Pythagorean theorem

h² = a² + b²

Since if you look at the rhombus it is formed by four right triangle we will take 1 of them with the following measures 5 cm (10/2) and height 12 cm (24/2) and replace:

h² = 5² + 12²

h² = 169

h = 13

now, the perimeter would be the sum of all its sides, which in this case are equal and measure 13 cm, therefore:

p = 4 * 13

p = 52

which means that the perimeter is 52 cm

Answer:

e=1o ajjajajajajkananwnwnwhhshssh

Complete Question:

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.

Answer:

a) P(exactly 5) = 0.209

b) P(at least six) = 0.183

c) P(less than four) = 0.358

Step-by-step explanation:

Sample size, n = 10

Proportion of adults that have very little confidence in newspapers, p = 41% p = 0.41

q = 1 - 0.41 = 0.59

This is a binomial distribution question:

[tex]P(X=r) = nCr p^{r} q^{n-r}[/tex]

a) P(exactly 5)

[tex]P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 252 * 0.01159 * 0.072\\P(X=5) = 0.209[/tex]

b) P(at least six)

[tex]P(X \geq 6) = P(6) + P(7) + P(8) + P(9) + P(10)[/tex]

[tex]P(X\geq6) = (10C6 * 0.41^6*0.59^4) + (10C7*0.41^7*0.59^3) + (10C8*0.41^8*0.59^2) + (10C9 *0.41^9*0.59^1) + (10C10 *0.41^{10})\\P(X\geq6) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001\\P(X\geq6) = 0.183[/tex]

c) P(less than four)

[tex]P(X < 4) = 1 - [x \geq 4][/tex]

[tex]P(X<4)= 1 - [P(4) + P(5) + P(x \geq 6)][/tex]

[tex]P(X <4)= 1 - [(10C4*0.41^4*0.59^6) + 0.209 + 0.183]\\P(X <4)= 0.358[/tex]

Answer:

Consider the following system of linear equations: 2 + 3y + 2z = 5 - 2x + y - z= -2 2x + 3z = 11 Instructions: • Solve the system by reducing its augmented matrix to reduced row echelon form (RREF). Yes, you must reduce it all the way to RREF. • Write out the matrix at each step of the procedure, and be specific as to what row operations you use in each step. • At the end of the procedure, clearly state the solution to the system outside of a matrix. • If the solution is unique, express the solution in real numbers. • If there are infinitely many solutions, express the solution in parameter(s). . If there is no solution, say so, and explain why.

All of the following are possible ranks of a 4x3 matrix EXCEPT O 1 2 3 4

How is the number of parameters in the general solution of a consistent linear system related to the rank of its coefficient matrix? Let r= number of rows in the coefficient matrix c= number of columns in the coefficient matrix p= number of parameters in the general solution R=rank of the coefficient matrix 1. R=p+r 2. R=C+p 3. R=r-p 4. R=C-p 5. R=p-r

Step-by-step explanation:

x + 3y +2z = 5

-2x + y - z = -2

2x + 3z = 11

Here,

[tex]A = \left[\begin{array}{ccc}1&3&2\\-2&1&-1\\2&0&3\end{array}\right][/tex]

[tex]B =\left[\begin{array}{ccc}5\\-2\\11\end{array}\right][/tex]

[tex]X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex]

i.e AX=B

We can write as augmented matrix

[tex]\left[\begin{array}{ccc|c}1&3&2&5\\-2&1&-1&-2\\2&0&3&11\end{array}\right][/tex]

[tex]\frac{R_2\rightarrow R_2+2R_1}{R_3\rightarrow R_3-2R_1} \left[\begin{array}{ccc|c}1&3&2&5\\0&7&3&8\\0&-6&-1&1\end{array}\right][/tex]

[tex]\frac{R_3\rightarrow R_3+\frac{6R_2}{7} }{R_1\rightarrow R_1-\frac{3R_2}{7} } \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&7&3&8\\0&0&11/7&55/7\end{array}\right][/tex]

[tex]\frac{R_2\rightarrow\frac{R_2}{7}}{R_3\rightarrow\frac{7}{11}R_3} \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&1&3/7&8/7\\0&0&11/7&55/7\end{array}\right][/tex]

[tex]\frac{R_1\rightarrow R_1 -\frac{5}{7}R_3}{R_2\rightarrow R_2 -\frac{3}{7}R_3} \left[\begin{array}{ccc|c}1&0&0&-2\\0&1&0&-1\\0&0&1&5\end{array}\right][/tex]

Since Rank (A|B) = Rank (A) = 3 = number of variables

Solution,

Length=5

Volume of cube=(L)^3

= 5*5*5

=125

hope it helps

Good luck on your assignment

Answer:

(a)                             n :      20           50          100         500

P (-200 < X - μ < 200) : 0.2886    0.4444    0.5954    0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable X represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, μ = $16,642 and standard deviation is, σ = $2,400.

Assuming that the random variable X follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{20}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{20}})[/tex]

                                           [tex]=P(-0.37<Z<0.37)\\\\=P(Z<0.37)-P(Z<-0.37)\\\\=0.6443-0.3557\\\\=0.2886[/tex]

[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{50}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{50}})[/tex]

                                           [tex]=P(-0.59<Z<0.59)\\\\=P(Z<0.59)-P(Z<-0.59)\\\\=0.7222-0.2778\\\\=0.4444[/tex]

[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{100}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{100}})[/tex]

                                           [tex]=P(-0.83<Z<0.83)\\\\=P(Z<0.83)-P(Z<-0.83)\\\\=0.7977-0.2023\\\\=0.5954[/tex]

[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{500}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{500}})[/tex]

                                           [tex]=P(-1.86<Z<1.86)\\\\=P(Z<1.86)-P(Z<-1.86)\\\\=0.9688-0.0312\\\\=0.9376[/tex]

                                 n :      20           50          100         500

P (-200 < X - μ < 200) : 0.2886    0.4444    0.5954    0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ([tex]\bar x[/tex]) approaches the whole population mean ([tex]\mu_{x}[/tex]).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

Using the normal distribution and the central limit theorem, it is found that:

a)

0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.

0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.

0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.

0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.

b)

b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Hence, the probability of sample of size n having a sample mean within 200 is the p-value of [tex]Z = \frac{200}{s}[/tex] subtracted by the p-value of [tex]Z = -\frac{200}{s}[/tex]

In this problem:

Item a:

For samples of 20, [tex]n = 20[/tex], and thus:

[tex]s = \frac{2400}{\sqrt{20}}[/tex]

Then

[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{20}}} = 0.37[/tex]

0.6443 - 0.3557 = 0.2886.

0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.

For samples of 50, [tex]n = 50[/tex], and thus:

[tex]s = \frac{2400}{\sqrt{50}}[/tex]

Then

[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{50}}} = 0.59[/tex]

0.7224 - 0.2776 = 0.4448.

0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.

For samples of 100, [tex]n = 100[/tex], and thus:

[tex]s = \frac{2400}{\sqrt{100}}[/tex]

Then

[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{100}}} = 0.83[/tex]

0.7967 - 0.2033 = 0.5934

0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.

For samples of 500, [tex]n = 500[/tex], and thus:

[tex]s = \frac{2400}{\sqrt{500}}[/tex]

Then

[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{500}}} = 1.86[/tex]

0.9686 - 0.0314 = 0.9372

0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.

Item b:

A larger sample size reduces the standard error, as [tex]s = \frac{\sigma}{\sqrt{n}}[/tex], that is, the standard error is inversely proportional to the square root of the sample size n, meaning that values are closer to the mean. Thus, the correct option is:

b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

A similar problem is given at https://brainly.com/question/24663213

Answer:

The standard deviation of the load distribution is of 5102.041 pounds.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

[tex]\mu = 20000[/tex]

Also, the probability that the load is between 10,000 and 30,000 pounds is 0.95.

10,000 pounds and 30,000 pounds are equidistant from the mean. Due to this, and the probability of 0.95 of having a value in this range, 10000 is the (100-95)/2 = 2.5th percentile and 30000 is the (100+95)/2 = 97.5th percentile. Applying one of them, we find the standard deviation.

30,000 is the 97.5th percentile:

This means that when X = 30000, Z has a pvalue of 0.975. So when X = 30000, Z = 1.96. Then

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.96 = \frac{30000 - 20000}{\sigma}[/tex]

[tex]1.96\sigma = 10000[/tex]

[tex]\sigma = \frac{10000}{1.96}[/tex]

[tex]\sigma = 5102.041[/tex]

The standard deviation of the load distribution is of 5102.041 pounds.

Answer:

[tex]= \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]

Step-by-step explanation:

Given:  3x1 - 6x2 + 9x3 = 0

x2 and x3 are free variables

We have:

3x1 = 6x2 - 9x3

divide all sides by 3, we have:

x1 = 2x2 - 3x3

Finding the general solution, we have:

[tex] \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{ccc}2x_2 - 3x_3\\x_2\\x_3\end{array}\right] [/tex]

[tex] = \left[\begin{array}{ccc}2x_2\\x_2\\0\end{array}\right] + \left[\begin{array}{ccc}-3x_3\\0\\x_3\end{array}\right][/tex]

[tex]= x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]

The general solution is

[tex]= \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]

Answer:

2,000 salmon

Step-by-step explanation:

The point estimate of the fraction of salmon that corresponds to the tagged Salmon released by the rangers, based on the fishermen report, is:

[tex]p = \frac{12}{80}\\p=0.15[/tex]

This means that the 300 salmon released into the lake correspond to 15% of the total salmon population. The estimate for the salmon population is:

[tex]P= \frac{300}{0.15}\\P=2,000\ salmon[/tex]

The estimate is 2,000 salmon.

Answer:

The estimate of the salmon population will be 2,000

Step-by-step explanation:

Rangers tagged and released 300 salmon into a Maine lake.

The fishermen reported catching 80 salmon of which 12 had tags

Thus we have 12/80 as a fraction of the tagged and released 300

 12/80 = 0.15 which is about 15% of the population of tagged and released 300.

To then estimate the salmon population, we have 300/0.15 = 2,000

Answer:

The probability of tossing a tail and then rolling a number greater than 5 is 0.188

Step-by-step explanation:

Independent events:

If two events, A and B, are independent, we have that:

[tex]P(A \cap B) = P(A)*P(B)[/tex]

In this question:

The coin and the die are independent. So

Event A: Tossing a tail.

Event B: Rolling a number greater than 5.

Probability of tossing a tail:

Coin can be heads or tails(2 outcomes), so the probability of a tail is [tex]P(A) = \frac{1}{2} = 0.5[/tex]

Probability of rolling a number greater than 5:

8 numbers(1 through 8), 3 of which(6,7,8) are greater than 5. So the probability of rolling a number greater than 5 is [tex]P(B) = \frac{3}{8} = 0.375[/tex]

Probability of A and B:

[tex]P(A \cap B) = P(A)*P(B) = 0.5*0.375 = 0.188[/tex]

The probability of tossing a tail and then rolling a number greater than 5 is 0.188

Answer:

no 8 is not divisible to 343060

Answer:

139.316148[tex]\pi[/tex]cubed inches

Step-by-step explanation:

If needed, use a calculator. Not sure if there's rounding for the final answer, so I will give you the precise answer.

1. Find the radius of the ball. If you know the circumference is 9.42 inches, divided it by 2, which the quotient is 4.71.

2. Remember the formula is V= [tex]\frac{4}{3} \pi r^3[/tex]

3. [tex]\frac{4}{3} \pi (4.71)^3[/tex]

4. [tex]\frac{4}{3} \pi (104.487111^3)[/tex]

5. [tex]4\pi (34.829037)[/tex]

6. 139.316148[tex]\pi[/tex]cubed inches

Answer:

2.5

Step-by-step explanation:

Answer:

2.75

Step-by-step explanation:

2 x 4=8 + 3=11/4=2.75

Answer:

The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute

P(x⁻>96) =0.0359

Step-by-step explanation:

Explanation:-

Given sample size 'n' =10

mean of the Population = 90 words per minute

standard deviation of the Population =10 wpm

we will use formula

                           [tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]

Let X⁻  = 96

                           [tex]Z = \frac{96-90 }{\frac{10}{\sqrt{10} } }[/tex]

                          Z =  1.898

The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute

[tex]P(X^{-}>x^{-} ) = P(Z > z^{-} )[/tex]

                   = 1- P( Z ≤z⁻)

                   = 1- P(Z<1.898)

                   = 1-(0.5 +A(1.898)

                   = 0.5 - A(1.898)

                   = 0.5 -0.4641 (From Normal table)

                  = 0.0359

Final answer:-

The probability that a random sample of 10 second grade students from  

                = 0.0359

Answer:

a) The third quartile Q₃ = 56.45

b) The variance = 2633.31

Step-by-step explanation:

a) The coefficient of skewness formula is given as follows;

[tex]SK = \dfrac{Q_{3}+Q_{1}-2Q_{_{2}}}{Q_{3}-Q_{1}}[/tex]

Plugging in the values, we have;

[tex]-0.38 = \dfrac{Q_{3}+30.8-2 \times 48.5_{_{}}}{Q_{3}-30.8}[/tex]

Solving gives Q₃ = 56.45

b) To determine the variance, we use the skewness formula as follows;

[tex]SK_{p} = \dfrac{Mean-\left (3\times Median - 2\times Mean \right )}{\sigma } = \dfrac{3\times\left ( Mean - Median \right )}{\sigma }[/tex]

Plugging in the values, we get;

[tex]-0.38= \dfrac{42-\left (3\times 48.5- 2\times 42\right )}{\sigma } = \dfrac{-19.5}{\sigma}[/tex]

[tex]\therefore \sigma =\dfrac{-19.5}{-0.38} = 51.32[/tex]

The variance = σ² = 51.32² = 2633.31.

Answer:

- $0.625

Step-by-step explanation:

To win 3 heads must be obtained, the probability of this is:

p = (1/2) ^ 3 = 0.125

Now, let's review the other scenarios:

HHH -> 0.125

HHT

THH ----> p (2H, 1T) = 3 * 0125 = 0.375

HTH

HTT

TTH ----> p (1H, 2T) = 3 * 0125 = 0.375

THT

TTT -> 0.125

So the waiting value would be:

EV = 2 * 0.125 - 1 * 0.375 - 1 * 0.375 - 1 * 0.125

EV = - 0.625

That is to say that the waiting value is - $ 0.625

The outcome og 1 game cannot be predicted but 100 you loss because the expected value is negative

The waiting value would be "-$0.625", the expected value will be negative so the outcome of game 1 can't be predicted but the 100 games you lose.

According to the question,

The probability will be:

→ [tex]p= (\frac{1}{2} )^3[/tex]

     [tex]= 0.125[/tex]

Now,

HHH:

→ 0.125

THH:

→ [tex]p(2H, 1T) = 3\times 0.125[/tex]

                   [tex]= 0.375[/tex]  

TTH:

→ [tex]p(1H, 2T) = 3\times 0.125[/tex]

                   [tex]= 0.375[/tex]

hence,

The waiting value will be:

→ [tex]EV = 2\times 0.125 -1\times 0.375-1\times -0.375-1\times 0.125[/tex]

         [tex]= -0.625[/tex]

Thus the above response is correct.

Learn more:

https://brainly.com/question/13753898

Answer:

For the sampling distribution of the sample proportion for a sample of size 50, the mean is 0.061 and the standard deviation is 0.034.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this question:

[tex]p = 0.061, n = 50[/tex]

So

Mean:

[tex]\mu = p = 0.061[/tex]

Standard deviation:

[tex]s = \sqrt{\frac{0.061*0.939}{50}} = 0.034[/tex]

For the sampling distribution of the sample proportion for a sample of size 50, the mean is 0.061 and the standard deviation is 0.034.