A box contains 11 red chips and 4 blue chips. We perform the following two-step experiment: (1) First, a chip is selected at random from the box and is then removed from the box. (After this first step, there are 14 chips left in the box. ) (2) Then, a chip is selected at random from the box (that is, from the remaining 14 chips) . Let B 1 be the event that the chip removed from the box at the first step of the experi- ment is red. Let B 2 be the event that the chip removed from the box at the first step of the experiment is blue. Let A be the event that the chip selected from the box at the second step of the experiment is red.Find P(B1), P(B2), P(A), P(B1|A), and P(B2|A).

Answer:

The longer diagonal has a length of 7.3 meters.

The angles are 31.65° and 18.35°

Step-by-step explanation:

If one angle of the parallelogram is 50°, another angle is also 50° and the other two angles are the supplement of this angle. so the other three angles are:

50°, 130° and 130°.

The longer diagonal will be the one opposite to the bigger angle (130°), and this diagonal divides the parallelogram in two triangles.

Using the law of cosines in one of these two triangles, we have:

[tex]diagonal^2 = a^2 + b^2 - 2ab*cos(130\°)[/tex]

[tex]diagonal^2 = 3^2 + 5^2 - 2*3*5*(-0.6428)[/tex]

[tex]diagonal^2 = 53.284[/tex]

[tex]diagonal = 7.3\ meters[/tex]

So the longer diagonal has a length of 7.3 meters.

To find the angles that this diagonal forms with the sides, we can use the law of sines:

[tex]a / sin(A) = b/sin(B)[/tex]

[tex]5 / sin(A) = diagonal / sin(130)[/tex]

[tex]sin(A) = 5 * sin(130) / 7.3[/tex]

[tex]sin(A) = 0.5247[/tex]

[tex]A = 31.65\°[/tex]

The other angle is B = 50 - 31.65 = 18.35°

Please check the image attached for better comprehension.

Answer:

The probability that demand during lead-time will exceed 20 bails is 0.2033.

Step-by-step explanation:

We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.

Let X = demand during the lead-time

So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                               Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu=[/tex] population mean demand = 15 bails

           [tex]\sigma[/tex] = standard deviation = 6 bails

Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)

       P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)

                                                             = 1 - 0.7967 = 0.2033

The image is missing, so i have attached it.

Answer:

x = 56.44°

Step-by-step explanation:

From the attached image, we can see that this is a right angle triangle which has opposite, adjacent and hypotenuse as sides. Since we want to find the angle x, thus, we can make use of trigonometric ratios.

From the attached image, the side opposite to angle x is 10ft and the hypotenuse is 12 ft.

From trigonometric ratios, we know that, sin x = opposite/hypotenuse

So, sin x = 10/12

x = sin^(-1) (10/12)

x = sin^(-1) 0.8333

x = 56.44°

Answer:

[tex]\boxed{ \ dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t\ }[/tex]

Step-by-step explanation:

it is a long time I have not applied Ito's lemma

I would say the following

for [tex]f(x)=x^2[/tex]

f'(x)=2x

f''(x)=2

so using Ito's lemma we can write that

[tex]dY_t=2V_tdV_t+\phi^2dt[/tex]

[tex]dY_t=2(\theta+\psi V_t^2)dt+2\phi V_tdW_t+\phi^2dt[/tex]

[tex]dY_t=(2\theta+2\psi V_t^2+\phi^2)dt+2\phi V_tdW_t[/tex]

so it comes

[tex]dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t[/tex]

Answer:

[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]

And replacing we got:

[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]

Step-by-step explanation:

For this case we have the following info given:

[tex] n=900[/tex] represent the sample size selected

[tex]p = 0.75[/tex] represent the population proportion

We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:

[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]

And the standard error is given;

[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]

And replacing we got:

[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]

Answer:

1/2 (simplified)

Step-by-step explanation:

6 numbers (that's the total probability) --> 6 denominator

3 are odd (odd numbers in the probability) --> 3 numerator

so => 3/6

--> simplify

1/2

Hope this helps!

Answer:

9.42 in²

Step-by-step explanation:

The area of whole circle S=pi*R²    , where pi is appr. 3.14,  R= 6 in

S= 3.14*6² =113.04 in²

The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²

The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.

To find the area of a sector, you can use the formula:

Area of sector = (θ/360) × π × r²

where θ is the central angle in degrees, r is the radius of the sector.

The central angle is 30 degrees and the radius is 6 inches.

Plugging these values into the formula:

Area of sector = (30/360) × π × 6²

= (1/12) × π × 36

= (1/12) × 3.14159 × 36

= 9.42478 square inches

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Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

Answer:

Step-by-step explanation:

The justification of each given statements in the question are:

11) F. Definition of right angle.

12) D. Definition of supplementary <'s.

13) A. Definition of congruence.

14) C. Definition of complementary <'s.

15) L. Congruent supplementary theorem

16) H. Vertical angle theorem.

17) G. Angle addition postulate.

18) J. Supplementary theorem.

Answer:

9.375 in^2

Step-by-step explanation:

Answer:

a) Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Step-by-step explanation:

a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:

Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Answer:

Step-by-step explanation:

[tex]H_0 : \texttt {null hypothesis}\\\\H_1 : \texttt {alternative hypothesis}[/tex]

The null hypothesis is the drink preferences are not changed at coffee shop.

The alternative hypothesis is the drink preferences are changed at coffee shop.

the level of significance = α = 0.05

We get the Test statistic

[tex]\texttt {Chi square}=\frac{\sum (F_o-F_e)}{F_e}[/tex]

Where, [tex]F_o[/tex] is observed frequencies and

[tex]F_e[/tex] is expected frequencies.

N = 6

Degrees of freedom = df = (N – 1)

= 6 – 1

= 5

the level of significance  α = 0.05

Critical value = 11.07049775

( using Chi square table or excel)

Tables for test statistic are given  below

                             F_o                F_e                 Chi square

Americanos           115                 153                9.4379

Capp.                     88                  94.5             0.447

Espresso               69                  63                 0.5714

Lattes                    59                  49.5               1.823

Macchiatos           44                   45                 0.022

Other                    75                   45                  20

Total                    450                 450                 32.30

[tex]\texttt {Chi square}=\frac{\sum (F_o-F_e)}{F_e}[/tex] = 32.30

P-value = 0.00000517

( using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

This is because their sufficient evidence to conclude that Drink preferences are changed at coffee shop.

Answer:

[tex] \: \: \: \: \: \: \: \: \: \: \dfrac{x}{5} = 7 \\ \implies \: x = 7 \times 5 \\ \implies \: x = 35[/tex]

So,b. multiplication

Answer:

A. division

Step-by-step explanation:

[tex]x/5=7[/tex]

[tex]x[/tex] is being divided by an integer.

[tex]x=35[/tex]

[tex]35/5=7[/tex]

35 divided by 5 is equal to 7.

Answer:

The degrees of freedom are given by:

[tex]df=n-1=317-1=316[/tex]

And replaicing we got:

[tex]29-2.1=26.9[/tex]    

[tex]29+2.1=31.1[/tex]    

The 95% confidence interval would be between 26.9 and 31.1

Step-by-step explanation:

Information given

[tex]\bar X= 29[/tex] represent the sample mean

[tex]\mu[/tex] population mean

s represent the sample standard deviation

[tex] ME= 2.1[/tex] represent the margin of error

n represent the sample size  

Solution

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

And this formula is equivalent to:

[tex] \bar X \pm ME[/te]x

The degrees of freedom are given by:

[tex]df=n-1=317-1=316[/tex]

And replaicing we got:

[tex]29-2.1=26.9[/tex]    

[tex]29+2.1=31.1[/tex]    

The 95% confidence interval would be between 26.9 and 31.1

The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .

The Laplace transform exist when s > 0 .

Here, the given function is f(t) = sin²(wt) .

The Laplace transform of the the function f(t),

F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }

F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }

F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }

F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]

Next,

The above Laplace transform exist if s > 0 .

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Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

Answer:

5/33649= approx 0.00015

Step-by-step explanation:

Total number of outcomes are  C24 6= 24!/(24-6)!/6!=19*20*21*22*23*24/(2*3*4*5*6)= 19*14*22*23

Half of the Committee =3 persons. That mens that number of the women in Commettee=3.  3 women from 6 can be elected C6  3  ways ( outputs)=

6!/3!/3!=4*5*6*/2/3=20

So the probability that 3 members of the commettee are women  is

P(women=3)= 20/(19*14*22*23)=5/(77*19*23)=5/33649=approx 0.00015

The probability that precisely half of the members will be women is;

P(3 women) = 0.1213

This question will be solved by hypergeometric distribution which has the formula;

P(x) = [S_C_s × (N - S)_C_(n - s)]/(NC_n)

where;

S is success from population

s is success from sample

N is population size

n is sample size

We are give;

s = 3 women (which is precisely half of the members selected)

S = 6 women

N = 24 men and women

n = 6 people selected

Thus;

P(3 women) = (⁶C₃ * ⁽¹⁸⁾C₍₃₎)/(²⁴C₆)

P(3 women) = (20 * 816)/134596

P(3 women) = 0.1213

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Answer:

114

Step-by-step explanation:

Answer:

144Step-by-step explanation:

Answer:

Height of tomato plant is the dependent variable

Presence of walnut leaves in the soil is the independent variable

Tomato plants grown without walnut leaves is the control

Step-by-step explanation:

An independent variable is the variable in an experiment that can be altered to test for a certain result. It is independent, or does not change with change in other factors in the experiment. In this case, the presence or absence, or quantity of walnut available in the soil is the independent variable in the experiment.

A dependent variable varies, and depends on the independent variable. It is what is measured in the experiment. In this case, the height of the tomato plants is the dependent variable that depends on the presence, absence or quantity of walnut in the soil.

A control in an experiment, is a replicate experiment, that is manipulated in order to be able to test a single variable at a time. Controls are variables are held constant so as to minimize their effect on the system under study. In this case, some of the tomato plants are planted without walnut in the soil, to test the effect of the absence of the walnut in the soil.

Answer:

r = 7.5

Step-by-step explanation:

Circle equation: [tex](x - h)^2 + (y - k)^2 = r^2[/tex]

Since we are already give r², we simply just take the square root of 56.25, and we should get 7.5 as our final answer!

Answer:

after the 1st year

Step-by-step explanation:

$2300 × 75% = $1725.00

$2300-$1725= $575

Answer:

It can travel 45 / 2 = 22.5 miles per gallon so the answer is 22.5 * 5.6 = 126 miles.

Answer:

2 and 256

Step-by-step explanation:

Check the attachment

Answer:

2  and255

Step-by-step explanation:

look atyourquestion

Answer: (fоg)(24)=5

Step-by-step explanation:

(fоg)(24) is f of g of 24. This means you plug in g(24) into f(x).

[tex]g(24)=\sqrt{24-8}[/tex]

[tex]g(24)=\sqrt{16}[/tex]

[tex]g(24)=4[/tex]

Now that we know g(24), we can plug it into f(x).

f(4)=2(4)-3

f(4)=8-3

f(4)=5

Answer:

a)[tex]A(t)=2000e^{0.085t}[/tex]

b)[tex]A'(t)=170e^{0.085t}[/tex]

c)$3059.1808

d)t=4.77 years

e) The balance growing is $254.99/year

Step-by-step explanation:

We are given that Two thousand dollars is deposited into a savings account at 8.5​% interest compounded continuously.

Principal = $2000

Rate of interest = 8.5%

a) What is the formula for​ A(t), the balance after t​ years? ​

Formula [tex]A(t)=Pe^{rt}[/tex]

So,[tex]A(t)=2000e^{0.085t}[/tex]

B)What differential equation is satisfied by​ A(t), the balance after t​ years?

So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]

[tex]A'(t)=170e^{0.085t}[/tex]

c)How much money will be in the account after 5 ​years? ​

Substitute t = 5 in the formula "

[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]

d)When will the balance reach ​$3000​?

Substitute A(t)=3000

So, [tex]3000=2000e^{0.085t}[/tex]

t=4.77

The balance reach $3000 in 4.77 years

e)How fast is the balance growing when it reaches ​$3000​?

Substitute the value of t = 4.77 in derivative formula :

[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]

Hence the balance growing is $254.99/year

Answer:

The complete factored form of this equation is (d + 6)²

Step-by-step explanation:

The first step in factoring this equation is multiply the first term and the last term together. Out first term is d² and our last term is 36. Since d² does not have a coefficient, then we assume this number to be 1.

1 × 36 = 36

So, now we need to find two factors that multiply to 36 and add together to get 12. Two factors that best represents this is 6 and 6. So, we will plug these numbers into our equation. Replace 12d with 6d + 6d.

d² + 6d + 6d + 36

Group the first two terms together and the last two terms together.

(d² + 6d) + (6d + 36)

Now, find the greatest common factor of each parentheses and factor the terms.

d(d + 6) + 6(d + 6)

From looking at this, we can tell that this equation is a perfect squared equation. So, this means instead of writing both parentheses, we can just write one of the parentheses and square it.

So, the factored form of this equation is (d + 6)²

Answer:

0.03125 = 3.125% probability that the person flipped 5 heads

Step-by-step explanation:

For each coin, there are only two possible outcomes. Either it was heads, or it was tails. The result of a coin toss is independent of other coin tosses. 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.

Five coins:

This means that n = 5.

Fair coin:

Equally as likely to be heads or tails, so p = 0.5.

What is the probability that the person flipped 5 heads?

This is P(X = 5).

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

[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125[/tex]

0.03125 = 3.125% probability that the person flipped 5 heads

Answer:

A. [tex](3b+8)^2[/tex]

Step-by-step explanation:

[tex]9b^2+48b +64\\=(3b)^2 + 2\times 3b\times 8 +(8)^2\\=(3b+8)^2[/tex]

Answer:

66.7%

Step-by-step explanation:

The numbers less than 7 on the list are 3, 4, 5, and 6.

4 numbers are less than 7 out of total 6 numbers.

4/6 = 2/3 = 0.667 = 66.7%

Answer:

slope = -4/5

Step-by-step explanation:

A line passes two points (x1, y1) and (x2, y2).

The slope of this line can be calculate by the formula:

s = (y2 - y1)/(x2 - x1)

=>The line that passes A(-4, 8) and B(-9, 12) has the slope:

s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5

Hope this helps!