Which of the following functions from R rightarrow R are one-to-one (injective), onto (surjective), or both (bijective)? Prove your answers. a. f(x) = 5x + 4 b. g(x) = 2x^2 - 2 c. h(x)= 1 + 2/x, x notequalto 0.

Answer:

[tex]P =0.3998[/tex]

Step-by-step explanation:

Let [tex]{\displaystyle {\overline{x}}}[/tex] be the average of the sample, and the population mean will be [tex]\mu[/tex]

We know that:

[tex]\mu = 4095[/tex] gr

Let [tex]\sigma[/tex] be the standard deviation and n the sample size, then we know that the standard error of the sample is:

[tex]E=\frac{\sigma}{\sqrt{n}}[/tex]

Where

[tex]\sigma=569[/tex]

[tex]n=130[/tex]

In this case we are looking for:

[tex]P(|{\displaystyle{\overline{x}}}- \mu|>42)[/tex]

This is:

[tex]{\displaystyle{\overline{x}}}- \mu>42[/tex] or [tex]{\displaystyle{\overline{x}}}- \mu<-42[/tex]

[tex]P=P({\displaystyle{\overline{x}}}- \mu>42)+ P({\displaystyle{\overline{x}}}- \mu<-42)[/tex]

Now we get the z score

[tex]Z=\frac{{\displaystyle{\overline{x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]P=P(z>\frac{42}{\frac{569}{\sqrt{130}}}) + P(z<-\frac{42}{\frac{569}{\sqrt{130}}})[/tex]

[tex]P=P(z>0.8416) + P(z<-0.8416)[/tex]

Looking at the tables for the standard nominal distribution we get

[tex]P =0.1999+0.1999[/tex]

[tex]P =0.3998[/tex]

Answer:

5.92 per cheesecake.

Step-by-step explanation:

The initial number cherry cheesecakes = 280,

Also, the cost of each cheesecakes = $ 2.51,

So, the total cost price = 280 × 2.51 = $ 702.8,

Markup = 65 %,

Thus, the total selling cost = 702.8 + 65% of 702.8 = $ 1159.62,

Now, the spoilage rate = 30 %,

So, the total new number of cheesecakes = 280 - 30% of 280 = 280 - 84 = 196,

Hence, the selling cost of each cheesecake = [tex]\frac{1159.62}{196}[/tex]

[tex]=\$ 5.91642857143[/tex]

[tex]\approx \$ 5.92[/tex]

Answer:

1) Solutions are x = 3 and x = 5/3

2) Solution are x ≤ 13/2 and  x ≤ -3/2

Step-by-step explanation:

1) Given absolute inequality,

|3x-7| = 2

⇒ 3x - 7 = ± 2

⇒ 3x = 7 ± 2

[tex]\implies x=\frac{7\pm 2}{3}[/tex]

[tex]x=\frac{7+2}{3}\text{ or }x=\frac{7-2}{3}[/tex]

[tex]\implies x = 3\text{ or }x=\frac{5}{3}[/tex]

2) l 2x-5 l ≤ 8

⇒  2x-5  ≤ ±8

⇒   2x ≤ 5 ± 8

[tex]\implies x\leq \frac{5\pm 8}{2}[/tex]

[tex]x\leq \frac{5+8}{2}\text{ or }x\leq \frac{5-8}{2}[/tex]

[tex]\implies x \leq \frac{13}{2}\text{ or }x\leq-\frac{3}{2}[/tex]

Answer:

Answer:

1) Solutions are x = 3 and x = 5/3

2) Solution are x ≤ 13/2 and  x ≤ -3/2

Step-by-step explanation:

1) Given absolute inequality,

|3x-7| = 2

⇒ 3x - 7 = ± 2

⇒ 3x = 7 ± 2

2) l 2x-5 l ≤ 8

⇒  2x-5  ≤ ±8

⇒   2x ≤ 5 ± 8

Answer:

Step-by-step explanation:

X= quarts of sauce # 1 ( 17% of sugar)

Y= quarts of sauce # 2( 30 % of sugar )

General Balance

[tex]X+Y=2.6[/tex]      EQ (1)

[tex]Y=2.6-X\\[/tex]

SUGAR BALANCE

[tex]0.17X+0.3Y=2.6*0.24\\[/tex]   EQ (2)

replace Y from eq(1) into eq(2)

[tex]0.17X+0.3(2.6-X)=0.624\\0.17X+0.78-0.3X=0.624\\-0.13X=0.624-0.780\\X=\frac{-0.156}{-0.13} \\X= 1.2\\Y=2.6-1.2=1.4[/tex]

Answer:

R indicates Random Assignment

Step-by-step explanation:

In designing an experiment, R is the symbol used for assigning the subjects randomly to different groups through randomization. It is called random assignment and It is the option (d) random assignment. It is achieved by using chance procedure ( random number generator, by flipping coin, throwing ). This ensures that all the subjects have equal chance of placing in all the groups.

Option (a) Symbol O is used for Observations. (So option (a) is wrong answer)

Option (b) Symbols x , y are used as Observational variables (So option (b) is wrong answer)

Option (c) Comparison groups is wrong answer as R is not used for that.

The symbol R in experimental design represents random assignment, a technique used to ensure that groups in an experiment are initially equivalent and thereby allow researchers to make causal inferences.

In experimental design, the symbol R indicates random assignment. Random assignment is a process used to create initial equivalence between the groups in an experiment, which is crucial for allowing researchers to draw causal conclusions. In this process, participants are assigned to different groups or conditions on a random basis, such as by drawing numbers or using a random number generator.

Random assignment minimizes the effects of lurking variables, ensuring that before the experimental manipulation occurs, the groups can be considered equivalent on various potential confounding factors. When individuals are selected into groups denoted by R, we can generally classify these designs as experimental, as opposed to nonexperimental designs where nonrandom assignment (NR) is indicated.

For example, in a study where the size of tableware is manipulated to see its effect on food consumption, the psychologist would use random assignment to ensure that both treatment groups are comparable prior to the intervention, which is the manipulation of the independent variable.

Proof:

Given any functions [tex]f_{1}(x),f_{2}(x),f_{3}(x)[/tex] they are linearly dependent if we can find values of [tex]c_{1},c_{2},c_{3}[/tex] such that

[tex]c_{1}f_{1}(x)+c_2f_{2}(x)+c_{3}f_{3}(x)=0[/tex]

Using the given functions in the above equation we get

[tex]c_{1}f_{1}(x)+c_2f_{2}(x)+c_{3}f_{3}(x)=0\\\\c_{1}x^{2}+c_{2}(1-x^{2})+c_{3}(2+x^{2})=0\\\\\Rightarrow (c_{1}-c_{2}+c_{3})x^{2}+c_{1}+c_{2}+2c_{3}=0[/tex]

This will be satisfied if and only if

[tex]c_1-c_2+c_3=0,c_1+c_2+2c_3=0[/tex]

Solving the equations we get

[tex]c_1+2c_3=-c_2\\\\c_1+c_1+2c_3+c_3=0\\2c_1+3c_3=0[/tex]

Since we have 3 variables and 2 equations thus we will get many solutions  

one being if we put [tex]c_3=1[/tex] we get

[tex]c_1+2c_3=-c_2\\\\c_1+c_1+2c_3+c_3=0\\2c_1+3c_3=0\\\\c_1=\frac{-3}{2}\\\\c_2=\frac{-1}{2}[/tex]

Thus we have [tex]c_1=\frac{-3}{2},c_2=\frac{-1}{2},c_3=1[/tex] as one solution. Hence the given functions are linearly dependent.

Answer:

Step-by-step explanation:

37

Answer:

131ft is the amount of fencing material needed

Step-by-step explanation:

Perimeter is the distance around a shape: we have to sum all the distances

P = d1 + d2 + d3

P = 34 ft + 30 ft + 67 ft = 131 ft

Answer:

D. 131 ft.

Step-by-step explanation:

If Sarah is planning to fence in her backyard garden and one side of the garden is 34 feet long, another side is 30 feet long, and the third side is 67 feet long. The perimeter of Sarah’s garden to determine the amount of fencing material needed is 131 feet.

Answer:

R = 2π×10⁻⁶ L / A

Step-by-step explanation:

Resistance varies directly with length and inversely with area, so:

R = kL/A

A round wire has cross-sectional area of:

A = πr²

Substituting:

R = kL/(πr²)

Given that R = 0.025 Ω when L = 500 cm and r = 0.2 cm:

0.025 = k (500) / (π (0.2²))

k = 2π×10⁻⁶

Therefore:

R = 2π×10⁻⁶ L / A

Answer:

0.00002 in the first blank. in the second blank its A

Step-by-step explanation:

Answer:

The peak fare price is $2,2

Step-by-step explanation:

Considering

Person 1  

5x + 4y= 17,40

Person 2

2x + 3y= 9,20

Isolating the x

2x= 9,20-3y

x=(9,20-3y)/2=4,6 -1,5y

x=4,6 -1,5y

Replacing in the other equation

5 (4,6 -1,5y) + 4y=17,40

23-7,5y+4y=17,40

23-17,40=7,5y-4y

5,6=3,5y

y=1,60 (this is the off peak fare)

x= 4,60-1,5y

x= 4,60-1,5*1,60

x=2,2 (this is the peak fare)

Final answer:

To determine the peak fare price, we formed a system of linear equations based on the given data. By using the elimination method, we solved the system to find that the peak fare price is $2.20.

Explanation:

We have a system of linear equations representing the total costs of peak and off-peak fares for two friends using public transportation in Washington D.C. Let's denote the peak fare as P and the off-peak fare as Op. According to the problem, we are given the following two equations based on the trips taken and the total cost paid by each person:

5P + 4Op = $17.40 (Person 1)

2P + 3Op = $9.20 (Person 2)

To find the peak fare price, we need to solve this system of equations. We can use a variety of methods, including substitution or the elimination method.

Step 1 - Use the Elimination Method:

Multiply the second equation by 2 to align the Op terms:

4P + 6Op = $18.40 (after multiplying by 2)

Step 2 - Subtract the resulting equation from the first person's total cost:

5P + 4Op - (4P + 6Op) = $17.40 - $18.40

P - 2Op = -$1.00

Step 3 - Solve for P:

Isolate P by adding 2Op to both sides:

P = 2Op - $1.00

Now, substitute P into the second equation from Person 2's total cost:

2(2Op - $1.00) + 3Op = $9.20

4Op - $2.00 + 3Op = $9.20

7Op - $2.00 = $9.20

Add $2.00 to both sides to get 7Op alone:

7Op = $11.20

Divide by 7 to find the off-peak fare:

Op = $11.20 / 7

Op = $1.60

Use the value of Op to find P:

P = 2($1.60) - $1.0

P = $3.20 - $1.00

P = $2.20

The peak fare price is $2.20.

Answer:

4/7

Step-by-step explanation:

To find the probability, we must first find the total number of golf balls. So 99+88+44 = 231 is our total. Now we must find the number of golf balls that are not a type A ball. This is just the total number of type B balls and type C balls. So 88+44 = 132. So the probability is 132/231. This simplifies to 12/21 or 4/7.

Answer:

Step-by-step explanation:

A) Suppose that we have the complex numbers

[tex]z= x + iy \quad \text{and} \quad \\\\ \tilde{z}=\tilde{x} + i \tilde{y}[/tex]

Remember that to sum complex numbers, we sum the real parts of the two numbers to get the real part and the imaginary parts of the two numbers to get the imaginary part. Hence,  

[tex]z+\tilde{z} = (x + i y) + (\tilde{x} + i \tilde{y}) = (x + \tilde{x})+i (y+\tilde{y})[/tex]

On the other hand, if we sum the matrix visualizations of [tex]z \quad \text{and} \quad \tilde{z}[/tex] we get

[tex]\left[\begin{array}{cc}x &y\\-y&x\end{array}\right] + \left[\begin{array}{cc}\tilde{x}&\tilde{y}\\ -\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x + \tilde{x}& y + \tilde{y}\\-(y+\tilde{y})&x+\tilde{x}\end{array}\right][/tex]

which is the matrix visualization of [tex]z + \tilde{z}[/tex].

To multiply two complex numbers, we use the distributive law to multiplly and then separete the real part from the imaginary part

[tex]z \cdot \tilde{z}= (x + iy) \cdot (\tilde{x} + i \tilde{y})=(x \tilde{x} + i x \tilde{y} + i \tilde{x} y - y\tilde{y} ) = (x\tilde{x}-y\yilde{y})+i(x\tilde{y}+\tilde{x}y)[/tex]

Again, if we multiply the matrix visualizations of [tex]z \quad \text{and} \quad \tilde{z}[/tex] we get

[tex]\left[\begin{array}{cc}x&y\\-y&x\end{array}\right]\left[\begin{array}{cc}\tilde{x}&\tilde{y}\\-\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x\tilde{x}-y\tilde{y}&x\tilde{y}+y\tilde{x}\\-y\tilde{x}-x\tilde{y}&x\tilde{x}-y\tilde{y}\end{array}\right][/tex]

which is the matrix viasualization of [tex]z\cdot\tilde{z}.[/tex]

B)  Since the usual matrix operations are consisten with the usual addition and multiplication rules in the complex numbers, we can use them to find the multiplicative inverses of a complex number [tex]z=x+iy[/tex].

We are looking for the complex number [tex]z^{-1}=(x+iy)^{-1}[/tex] which in terms of matrices is equivalent to find the matrix

[tex]\left[\begin{array}{cc}x&y\\-y & x\end{array}\right]^{-1}= \dfrac{1}{x^{2}+y^{2}} \left[\begin{array}{ccc}x&-y\\y&x\end{array}\right][/tex]    

Hence,

[tex]z^{-1}=\dfrac{1}{x^2 +y^2} (x-iy)=\dfrac{1}{|z|^2}(x-iy)[/tex]

Answer:

Intersection: {w, 7, y}

Union: {a, m, w, u, 7, y, g}

Step-by-step explanation:

The intersection of two or more sets are the elements they have in common, that means the elements that are in all the sets at the same time.

For example, "a" is in A but not in B, that's why is not in the intersection. On the other hand, "w" is in both sets, so it's in the intersection.

The union are all the elements that are in one set or the other, but we don't add the element twice if it's in both sets.  

For example, "a" is in A so we add it to the union. "w" is in A so we add it to the union but it's also in B, we don't add it again because it is already in the union.  

Answer:

The bill will be $26070.

Step-by-step explanation:

Your base plan costs you $30 a month

And each text sent or received is only 12 cents or $0.12.

Let the messages sent be = x

So, cost for messages become = [tex]0.12x[/tex]

Let total bill cost be = y

Now, total bill will comprise of monthly plan charges plus message charges.

So, equation forms:

[tex]y=30+0.12x[/tex]

Now, given is that In one month 217,000 messages were sent between two brothers in Philadelphia.

So, their approximate bill will be :

[tex]y=30+0.12(217000)[/tex]

=>[tex]y=30+26040[/tex]

=> y = 26070

Therefore, bill will be $26070.

The answer explains how to write an equation for the total monthly cost of a phone plan based on text message usage and calculates the bill for 217,000 messages sent between two brothers in Philadelphia.

To write an equation for the total monthly cost of a phone plan with a base cost and a cost per text message:

For the specific case of 217,000 messages sent between two brothers:

Answer: Approximately $18,000

Step-by-step explanation:

Since you are approximating, the 7 indicates the need to round the original number up to 36,000. Then divide by 2 since you want half, thus getting $18,000.

Answer:

210

Step-by-step explanation:

In 2005 the employee was tripple of 1995 which is 50×3=150. So number of employees in 2005=150.

In 2015 number of employees were 40% of employees in 2005.

40% of 150 =150×40÷100 = 60,

Therefore number of employees in 2015 =150+60= 210 employees.

Answer:

210 employees in 2015

Step-by-step explanation:

In order to solve this we first have to calculate the number of employees that there were in 2005, since the problem says that they tripled their employees from 1995, that meas three times 50, meaning 150 employees in 2005, if by 2015 the number of employees had increase in 40% we just do a simple rule of three, 150 being the 100% and trying to calculate the 140%:

[tex]\frac{150}{100}= \frac{x}{15} \\x=\frac{150*15}{100}\\ x= 210[/tex]

So now we know that there were 210 employees in 2015.

Answer:  0.4404

Step-by-step explanation:

Let the random variable x is normally distributed .

Given : Mean : [tex]\mu=\ 87[/tex]

Standard deviation : [tex]\sigma= 55[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = 79

[tex]z=\dfrac{79-87}{55}\approx-0.15[/tex]

The p-value = [tex]P(x<79)=P(z<-0.15)[/tex]

[tex]=0.4403823\approx0.4404[/tex]

Hence, the required probability : [tex]P(x<79)=0.4404[/tex]

a)

[tex]6\cdot5\cdot2\cdot2=120[/tex]

b)

[tex]4\cdot5\cdot2\cdot2 +2\cdot5\cdot1\cdot1=80+10=90[/tex]

Answer: 0.0081

Step-by-step explanation:

Let X be the number of rafts.

Given : The mean number of rafts floating : [tex]\lambda=0.4[/tex] rafts per day .

Then , for 7 days the number of rafts = [tex]\lambda_1=\lambda\times 7=0.4\times7=2.8[/tex] rafts per day .

The formula to calculate the Poisson distribution is given by :_

[tex]P(X=x)=\dfrac{e^{-\lambda_1}\lambda_1^x}{x!}[/tex]

Now, the  probability that they will have to wait more than a week is given by :-

[tex]P(X>7)=1-P(X\leq7)\\\\=1-(P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(5)+P(6)+P(7))\\\\=1-(\dfrac{e^{-2.8}2.8^{0}}{0!}+\dfrac{e^{-2.8}2.8^{1}}{1!}+\dfrac{e^{-2.8}2.8^{2}}{2!}+\dfrac{e^{-2.8}2.8^{3}}{3!}+\dfrac{e^{-2.8}2.8^{4}}{4!}+\dfrac{e^{-2.8}2.8^{5}}{5!}+\dfrac{e^{-2.8}2.8^{6}}{6!}+\dfrac{e^{-2.8}2.8^{7}}{7!})\\\\=1-0.991869258012=0.008130741988\approx0.0081[/tex]

Hence, the required probability : 0.0081

The question asks for the probability that the sample mean of income differs from the true mean by over 42 dollars for a sample size of 412. We use the Central Limit Theorem to approach this problem, calculating the standard error and Z-score to refer to the standard normal distribution to find the associated probability.

This question pertains to the field of statistics, more specifically, the Central Limit Theorem. The Central Limit Theorem says that if we take many samples from a population, the distribution of the sample means will approximate a normal distribution, regardless of the shape of the population distribution. We can use this theorem to calculate a probability related to the sample mean.

In this case, the population mean (μ) is a per capita income of 24,787 dollars and the population variance (σ²) is 169,744 dollars. We're asked to find the probability that the sample mean would differ from the true mean by more than 42 dollars if a sample of 412 persons is randomly selected.

The standard error of the sample mean is calculated by σ / sqrt(n), where σ is the standard deviation (sqrt(σ²)), and n is the sample size (412). After finding the standard error, we will calculate the Z-score of 42, which is the number of standard errors 42 is away from the mean. Calculating the Z-score is achieved by z = (X - μ) / SE, where X is the value of 42.

We can refer the calculated Z-score to the standard normal distribution to find the associated probability. However, since we are looking for the probability of a difference greater than 42, we want the probabilities in the tails of the distribution beyond our calculated Z-score, so it would be 1 minus this value.

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Answer:  The required value of the given expression is (h +1).

Step-by-step explanation:  We are given a function f(x) as follows :

[tex]f(x)=x^2-3x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We are to evaluate the following :

[tex]E=\dfrac{f(2+h)-f(2)}{h}.[/tex]

Substituting x = 2 + h in equation (i), we get

[tex]f(2+h)\\\\=(2+h)^2-3(2+h)\\\\=(4+4h+h^2)-6-3h\\\\=4+4h+h^2-6-3h\\\\=h^2+h-2[/tex]

and substituting x = 2 in equation (i), we get

[tex]f(2)=2^2-3\times2=4-6=-2.[/tex]

Therefore, the value of expression E can be evaluated as follows :

[tex]E\\\\\\=\dfrac{f(2+h)-f(2)}{h}\\\\\\=\dfrac{(h^2+h-2)-(-2)}{h}\\\\\\=\dfrac{h^2+h-2+2}{h}\\\\\\=\dfrac{h^2+h}{h}\\\\\\=\dfrac{h(h+1)}{h}\\\\=h+1.[/tex]

Thus, the required value of the given expression is (h +1).

Answer:

r(q(-4)) = -10

Step-by-step explanation:

q(x) = -x - 2

r(x) = -2x^2 -2

r(q(-4))

First find q(-4)

Let x=-4

q(-4) = -(-4) - 2

q(-4) = +4 -2 = 2

q(-4) =2

We substitute this value in for x in r(x)

r(2) = -2(2)^2 -2

     = -2 (4) -2

     = -8 -2

     = -10

For this case we have the following functions:

[tex]q (x) = - x-2\\r (x) = - 2x ^ 2-2[/tex]

We must find [tex]r (q (x))[/tex]. So:

We substitute [tex]q (x)[/tex]in [tex]r (x).[/tex]

[tex]r (q (x)) = - 2 (-x-2) ^ 2-2[/tex]

Now we substitute[tex]x = -4[/tex]

[tex]r (q (-4)) = - 2 (- (- 4) -2) ^ 2-2\\r (q (-4)) = - 2 (+ 4-2) ^ 2-2\\r (q (-4)) = - 2 (2) ^ 2-2\\r (q (-4)) = - 2 (4) -2\\r (q (-4)) = - 8-2\\r (q (-4)) = - 10[/tex]

Answer:

[tex]r (q (-4)) = - 10[/tex]

Answer:

450 students

Step-by-step explanation:

Let s = number of students

1/2 are Americans  1/2s are Americans

s -1/2s = number of students left

1/2 s = number of students left

1/3 of the students left are Europeans

1/3 (1/2s) = 1/6s  are Europeans

The rest are Australians = 150

Students = Americans + Europeans + Australians

   s          =   1/2s          +    1/6s    + 150

Combine like terms

s    = 1/2*3/3 *s + 1/6s +150

s = 3/6s + 1/6s +150

s = 4/6s + 150

Subtract 4/6s from each side

s -4/6s = 150

6/6s - 4/6s = 150

2/6s = 150

1/3s = 150

Multiply each side by 3

3* 1/3s = 150*3

s = 450

There are 450 students

Answer:

There are 450 students.

Step-by-step explanation:

Let's call

s = total number of students

am = American students

eu = European students

au = Australian students

Half of the students in an International School are Americans. Then,

1/2 s = am [1]

One third of the remaining students are Europeans.

The remaining students are s - 1/2 s = 1/2 s

1/3 × (1/2 s) = 1/6 s = eu  [2]

There are 150 Australians.

au = 150 [3]

The total number of students is the sum of Americans, Europeans and Australians.

am + eu + au = s

Replacing [1], [2] and [3].

1/2 s + 1/6 s + 150 = s

2/3 s + 150 = s

1/3 s = 150

s = 450

Answer:

r=5 units

Step-by-step explanation:

we have

[tex]x^{2}+y^{2} -4y=21[/tex]

Convert the equation of the circle in standard form

Complete the square twice. Remember to balance the equation by adding the same constants to each side

[tex]x^{2}+(y^{2} -4y+4)=21+4[/tex]

[tex]x^{2}+(y^{2} -4y+4)=25[/tex]

Rewrite as perfect squares

[tex]x^{2}+(y-2)^{2}=25[/tex]

center (0,2)

radius r=5 units

Answer:

The probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement is 0.03125 .

The probability that the first two cards chosen are spades and the third card is red if the cards are chosen without replacement is 0.03058

Step-by-step explanation:

Total no. of cards = 52

Spade cards = 13 (Black)

Club cards = 13 (Black)

Heart cards = 13 (Red)

Diamond cards = 13(Red)

Total red cards = 26

With replacement case:

Probability of getting spade on first draw = [tex]\frac{13}{52}[/tex]

Now card is replaced

Probability of getting spade on second draw = [tex]\frac{13}{52}[/tex]

Now card is replaced

Probability of getting red card on third draw = [tex]\frac{26}{52}[/tex]

So, The probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement =  [tex]\frac{13}{52} \times \frac{13}{52} \times \frac{26}{52}[/tex]

                                                                                        =[tex]\frac{1}{32}[/tex]

                                                                                        =  [tex]0.03125[/tex]

Without replacement case:

Probability of getting spade on first draw = [tex]\frac{13}{52}[/tex]

Remaining cards = 51

Remaining spade cards = 12

Probability of getting spade on second draw = [tex]\frac{12}{51}[/tex]

Remaining cards = 50

Probability of getting red card on third draw = [tex]\frac{26}{50}[/tex]

So, The probability that the first two cards chosen are spades and the third card is red if the cards are chosen without replacement =  [tex]\frac{13}{52} \times \frac{12}{51} \times \frac{26}{50}[/tex]

                                                                                       = [tex]\frac{13}{425}[/tex]

                                                                                        =  [tex]0.03058[/tex]

Hence The probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement is 0.03125 . The probability that the first two cards chosen are spades and the third card is red if the cards are chosen without replacement is 0.03058

When the cards are chosen with replacement, the probability is 169/2704. When the cards are chosen without replacement, the probability is 39/884.

When the cards are chosen with replacement, the probability that the first two cards chosen are spades and the third card is red can be found by multiplying the probabilities of each event happening. There are 13 spades in a deck of 52 cards, so the probability of choosing a spade on the first draw is 13/52. Since the cards are chosen with replacement, the probability of choosing a spade on the second draw is also 13/52. The probability of choosing a red card on the third draw is 26/52, since there are 26 red cards in the deck. Therefore, the overall probability is (13/52) * (13/52) * (26/52) = 169/2704.

When the cards are chosen without replacement, the probability changes. The probability of choosing a spade on the first draw is 13/52. Since the first card is not replaced, there are now 51 cards left and 12 spades remaining. So the probability of choosing a spade on the second draw is 12/51. Finally, there are 25 red cards left out of 50 total cards, so the probability of choosing a red card on the third draw is 25/50. Therefore, the overall probability is (13/52) * (12/51) * (25/50) = 39/884.

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Answer: the technicians have 19 hours to correct the problem.

Step-by-step explanation:

The countdown started at "T-minus 27 hours" that means that the clock starts ticking backwards and if an hour passes away the lime left for the launch will be 26 hours and the limit the technicians have is T minus 8 hours so the problem can be solved with a simple substraction 27 hours minus 8 hours equals 19 hours.

Answer:

wow that school has alot of test don't it lol

Step-by-step explanation:

anyway what's the question

After calculating the Z-scores for the student's performance in each class, we found that she performed closer to the top of her class in statistics. Thus, she did relatively better on the statistics exam compared to the chemistry exam.

To determine relative performance in each class, we need to calculate the number of standard deviations the student's score is from the mean (also known as a

Z-score

).

For the chemistry exam, where the mean was 90 and the standard deviation was 64, the student's Z-score is: (102 - 90) / 64 = 0.1875.

For the statistics exam, with a mean of 70 and standard deviation of 16, the student's Z-score is: (77 - 70) / 16 = 0.4375.

Comparing these Z-scores, we can conclude that the student did relatively better in the statistics class because her Z-score (0.4375) was higher there than in her chemistry class (0.1875). This indicates she performed closer to the top of her class in statistics. Therefore, the answer would be selection C: the student did relatively better on the statistics exam than on the chemistry exam, compared to the other students in the two classes.

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The probability that no more than [tex]25[/tex] were victims of e-mail fraud is [tex]\fbox{0.0278}[/tex].

Further explanation:

Given:

The probability [tex]p[/tex] that an appliance is currently repaired is [tex]0.5[/tex].

The number of complex [tex]n[/tex] are [tex]100[/tex].

Calculation:

The [tex]\bar{X}[/tex] follow the Binomial distribution can be expressed as,

[tex]\bar{X}\sim \text{Binomial}(n,p)[/tex]

Use the normal approximation for [tex]\bar{X}[/tex] as

[tex]\bar{X}\sim \text{Normal}(np,np(1-p))[/tex]

The mean [tex]\mu[/tex] is [tex]\fbox{np}[/tex]

The standard deviation [tex]\sigma\text{ } \text{is} \text{ } \fbox{\begin{minispace}\\ \sqrt{np(1-p)}\end{minispace}}[/tex]

The value of [tex]\mu[/tex] can  be calculated as,

[tex]\mu=np\\ \mu= 100 \times0.5\\ \mu=50[/tex]

The value of [tex]\sigma[/tex] can be calculated as,

[tex]\sigma=\sqrt{100\times0.5\times(1-0.5)} \\\sigma=\sqrt{100\times0.5\times0.5}\\\sigma=\sqrt{25}\\\sigma={5}[/tex]

By Normal approximation \bar{X} also follow Normal distribution as,

[tex]\bar{X}\sim \text{Normal}(\mu,\sigma^{2} )[/tex]

Substitute [tex]50[/tex] for [tex]\mu[/tex] and [tex]25[/tex] for [tex]\sigma^{2}[/tex]

[tex]\bar{X}\sim\text {Normal}(50,25)[/tex]

The probability that at least [tex]60[/tex] are currently being repaired can  be calculated as,

[tex]\text{Probability}=P\left(\bar{X}>60)\right}\\\text{Probability}=P\left(\frac{{\bar{X}-\mu}}{\sigma}>\frac{{(60-0.5)-50}}{\sqrt{25} }\right)\\\text{Probability}=P\left(Z}>\frac{{59.5-50}}{5}}\right)\\\text{Probability}=P\left(Z}>\frac{9.5}{5}}\right)\\\text{Probability}=P(Z}>1.9})[/tex]

The Normal distribution is symmetric.

Therefore, the probability of greater than [tex]1.9[/tex] is equal to the probability of less than [tex]1.9 [/tex].

[tex]P(Z>1.9})=1-P(Z<1.9)\\P(Z>1.9})=1-0.9722\\P(Z>1.9})=0.0278[/tex]

Hence, the probability that no more than [tex]25[/tex] were victims of e-mail fraud is [tex]\fbox{0.0278}[/tex].

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Answer Details:

Grade: College Statistics

Subject: Mathematics

Chapter: Probability and Statistics

Keywords:

Probability, Statistics, Appliance, Apartment complex, Binomial distribution, Normal distribution, Normal approximation, Central Limit Theorem, Z-table, Mean, Standard deviation, Symmetric.

Answer:

1. The probability that the student will get exactly 6 correct answers is [tex]\frac{1120}{19683}[/tex].

2. The probability that the student will get more than 6 correct answers is [tex]\frac{43}{2187}[/tex].

Step-by-step explanation:

From the given information it is clear that

The total number of equations (n) = 10

The probability of selecting the correct answer (p)= [tex]\frac{1}{3}[/tex]

The probability of selecting the incorrect answer (q)= [tex]1-p=1-\frac{1}{3}=\frac{2}{3}[/tex]

According to the binomial distribution, the probability of selecting r items from n items is

[tex]P=^nC_rp^rq^{n-r}[/tex]

where, p is probability of success and q is the probability of failure.

The probability that the student will get exactly 6 correct answers is

[tex]P(r=6)=^{10}C_6(\frac{1}{3})^6(\frac{2}{3})^{10-6}[/tex]

[tex]P(r=6)=210(\frac{1}{3})^6(\frac{2}{3})^{4}=\frac{1120}{19683}[/tex]

Therefore the probability that the student will get exactly 6 correct answers is [tex]\frac{1120}{19683}[/tex].

The probability that the student will get more than 6 correct answers is

[tex]P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{10-7}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{10-8}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{10-9}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{10-10}[/tex]

[tex]P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{3}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{2}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{1}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{0}[/tex]

[tex]P(r>6)=120\times \frac{8}{59049}+45\times \frac{4}{59049}+10\times \frac{2}{59049}+1\times \frac{1}{59049}=\frac{43}{2187}[/tex]

Therefore the probability that the student will get more than 6 correct answers is [tex]\frac{43}{2187}[/tex].