Bryson hopes to win a three-day vacation in a drawing that is being held at his office. He purchased 40 raffle tickets. There were 500 raffle tickets sold. What is the theoretical probability of Bryson winning the trip?

Answer:

  (x, y) = (-4, -6)

Step-by-step explanation:

See the attachment for a graph. The solution is the point of intersection of the two lines.

Answer: (-4, -6)

Step-by-step explanation: If you go to the graphing calculator on edge, click the middle part where both lines intersect and it should tell you the pairs). :)

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:

None of the choices are correct

Step-by-step explanation:

As you can see in the figures, the side lengths of ABC are twice the length of the corresponding sides of MNL. So the scale factor from ABC to MNL would 1/2, which is not any of the options.

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:

  $6534.42

Step-by-step explanation:

Put the given values into the simple interest formula and solve for the remaining variable.

  A = P(1 +rt)

where P is the principal invested, r is the annual rate, and t is the number of years.

  $7000 = P(1 +0.095(9/12)) = 1.07125P

  $7000/1.07125 = P ≈ $6534.42

The value that must be invested is $6534.42.

Answer:

His weekly exercise time goal is 168 minutes.

Step-by-step explanation:

This question can be solved using a rule of three.

42 minutes is 25% = 0.25 of the total

x minutes is 100% = 1 of the total.

Then

42 minutes - 0.25

x minutes - 1

[tex]0.25x = 42[/tex]

[tex]x = \frac{42}{0.25}[/tex]

[tex]x =  168[/tex]

His weekly exercise time goal is 168 minutes.

Answer:

x = 3.4

Step-by-step explanation:

Step 1: Isolate x

0.5x = 1.7

Step 2: Divide both sides by 0.5

x = 3.4

And we have our final answer!

Answer: x=3.4

Step-by-step explanation:

[tex]0.5x+4.2=5.9[/tex]

multiply both sides by 10

[tex]5x+42=59[/tex]

subtract 42 on both sides

[tex]5x=17[/tex]

divide 5 on both sides

[tex]x=\frac{17}{5}[/tex] or

Simplify

x= 3.4

Answer:

2.86=28 tenths+6 hundredths.

Step-by-step explanation:

2.86=2 ones+8 tenths+6 hundredths.

2.86=28 tenths+6 hundredths.

Answer:

see

Step-by-step explanation:

ones .  tenths  hundredths

2      .     8          6

8 tenths

If we are not using the ones place

we have 28 tenths

Answer:

The statistical test to be used is the paired t-test.

Step-by-step explanation:

The dependent t-test (also known as the paired t-test or paired-samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference between these two means.

We use the paired t-test if we have two measurements on the same item, person or thing. We should also use this test if we have two items that are being measured with a unique condition.

For instance, an experimenter tests the effect of a medicine on a group of patients before and after giving the doses.

Similarly, in this case a paired t-test would be used to deter whether there was any changes in the cholesterol level within each group as result of the treatment.

Thus, the statistical test to be used is the paired t-test.

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]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:

9.375 in^2

Step-by-step explanation:

Answer:

A' - (8,2)

B' - (5,1)

C' - (4,2)

D' - (4,5)

Step-by-step explanation:

See attachment for the missing figure.

We can see that the vertices of the polygon ABCD have coordinates A(4,6), B(5,3), C(4,2) and D(1,2)

Polygon ABCD is rotated 90° clockwise about point C to form polygon A′B′C′D′ (see attached diagram), then

A'(8,2);

B'(5,1);

C' is the same as C, thus, C'(4,2);

D'(4,5).

Answer:

tiStep-by-step explanaon:

Answer:

a) [tex] E(Y) = \sum_{i=1}^n Y_i P(Y_i)[/tex]

And replacing we got:

[tex] E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60[/tex]

b) [tex] E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1[/tex]

And then the expected value would be:

[tex] E(100Y^2) = 100*1.1= 110[/tex]

Step-by-step explanation:

We assume the following distribution given:

Y       0       1        2        3

P(Y) 0.60 0.25  0.10  0.05

Part a

We can find the expected value with this formula:

[tex] E(Y) = \sum_{i=1}^n Y_i P(Y_i)[/tex]

And replacing we got:

[tex] E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60[/tex]

Part b

If we want to find the expected value of [tex] 100 Y^2[/tex] we need to find the expected value of Y^2 and we have:

[tex] E(Y^2) = \sum_{i=1}^n Y^2_i P(Y_i)[/tex]

And replacing we got:

[tex] E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1[/tex]

And then the expected value would be:

[tex] E(100Y^2) = 100*1.1= 110[/tex]

*see attachment showing the 2 boxes

Answer:

3 ft²

Step-by-step explanation:

==>Given:

Box J with the following dimensions:

L = 4.5ft

W = 3ft

H = 2ft

Box F:

L = 3ft

W = 3ft

H = 3ft

==>Required:

Difference between the surface area of box J and box F

==>Solution:

Surface area = 2(WL + HL + HW)

=>S.A of box J = 2(3*4.5 + 2*4.5 + 2*3)

= 2(13.5 + 9 + 6)

= 2(28.5)

S.A of box J = 57 ft²

=>S.A of box F = 2(3*3 + 3*3 + 3*3)

= 2(9 + 9 + 9)

= 2(27)

S.A of box F = 54 ft²

Difference between box J and box F = 57 - 54 = 3 ft²

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:

The amount of dividend Giant Machinery can  pay its shareholders this year =   $136,250

The Dividend Payout ratio of the company = 54.50%

Ex-dividend price = $22.875

The current value of the firm's equity in total  = $9196428.571

The current value of the firm's equity per share = $6.131

Step-by-step explanation:

Given that:

The Current Capital Structure = 65% equity and 35% debt.

The net income in the current year = $250,000

Also, the company is planning to launch a project that will requires an investment of  $175,000 next year.

The current share price = $25/share

(a)

The first objective is to determine how much dividend Giant Machinery can pay its shareholders this year and what is dividend payout ratio of the company.

The amount of dividend Giant Machinery can  pay its shareholders this year= Net profit - Investment amount × (percentage of equity)

= 250,000 - 175,000 × (65%)

= 250,000 - 113,750

= $136,250

The dividend payout ratio of the company.can be calculated as follows:

Dividend Payout ratio of the company [tex]= \mathbf{\dfrac{ total \ \ dividends}{total \ \ earning}}[/tex]

We all know that the net income in the current year is the  Total Earning which is 250,000

Thus;

The Dividend Payout ratio of the company [tex]= \mathbf{\dfrac{ 136250}{250000}}[/tex]

The Dividend Payout ratio of the company = 0.545

The Dividend Payout ratio of the company = 54.50%

b).

If the company is paying a dividend of $2.50/share and tomorrow the stock will go ex-dividend.

Also, assuming:

the tax on dividend =  15%.

We are to calculate the ex-dividend price tomorrow morning.

To calculate the ex-dividend price tomorrow morning; we use the relation:

Ex-dividend price which is the result of the difference between the current price and dividend multiply by the difference between the 1 and the tax on the dividend

i.e

Ex-dividend pric = current price - Dividend × (1 - tax on dividend)

Given that :

The current share price = $25/share

Therefore;

Ex-dividend price = $25 - $2.5 × ( 1 - 15%)

Ex-dividend price = $25 - $2.5 × ( 1 - 0.15)

Ex-dividend price = $25 - $2.5  × 0.85

Ex-dividend price = $25 - $2.125

Ex-dividend price = $22.875

C)

From the  information given in the part C of the question:

the company now plans to pays a total dividend of $2.5 million.= $ 2,500.000

Let say the year the dividend was paid was 0 year

and 7.5 million one year from now as a liquidating dividend.

So; the dividend paid in year 1 now = 7.5 million = $ 7,500,000

The required rate of return for shareholders is 12%

Outstanding share of the firm = 1.5 million = $1,500,000

We are to calculate  the current value of the firm’s equity in total and per share if the firm has 1.5 million shares outstanding.

To start with the current value of the firm’s equity in total;

The current value of the firm's equity in total =[tex]\mathbf{amount \ of \ dividend \ paid \ in \ 0 \ year + \dfrac {amount \ of \ dividend \ paid \ in \ 1 \ year } { 1+required \ rate \ of \ return } }[/tex]= [tex]\mathbf{2,500,000 + \dfrac{7,500,000 }{1+ 0.12} }[/tex]

[tex]=\mathbf{2,500,000 + \dfrac{7,500,000 }{1.12} }[/tex]

= 2,500,000 + 6696428.571

= $ 9196428.571

The current value of the firm's equity per share =[tex]\mathbf{ \dfrac{ current \ value \ of \ the \ firm's \ equity \ in \ total}{ Outstanding \ share }}[/tex]

[tex]\mathbf{= \dfrac{9196428.571}{1500000}}[/tex]

= $6.130952381

≅ $6.131

The current value of the firm's equity per share = $6.131

Answer:

[tex]A.\ y = x^2 - 6x + 13[/tex] is the correct answer.

Step-by-step explanation:

We know that vertex equation of a parabola is given as:

[tex]y = a(x-h)^2+k[/tex]

where [tex](h,k)[/tex] is the vertex of the parabola and

[tex](x,y)[/tex] are the coordinate of points on parabola.

As per the question statement:

The parabola opens upwards that means coefficient of [tex]x^{2}[/tex] is positive.

Let [tex]a = +1[/tex]

Minimum of parabola is at x = 3.

The vertex is at the minimum point of a parabola that opens upwards.

[tex]\therefore[/tex] [tex]h = 3[/tex]

Putting value of a and h in the equation:

[tex]y = 1(x-3)^2+k\\\Rightarrow y = (x-3)^2+k\\\Rightarrow y = x^2-6x+9+k[/tex]

Formula used: [tex](a-b)^2=a^{2} +b^{2} -2\times a \times b[/tex]

Comparing the equation formulated above with the options given we can observe that the equation formulated above is most similar to option A.

Comparing [tex]y = x^2 - 6x + 13[/tex] and [tex]y = x^2-6x+9+k[/tex]

13 = 9+k

k = 4

Please refer to the graph attached.

Hence, correct option is [tex]A.\ y = x^2 - 6x + 13[/tex]

Answer:

A. y = x^2 -6x + 13

Step-by-step explanation:

Answer:

x= 8.00 Interest rate on $14000

y= 7.50 Interest rate on $7000

Step-by-step explanation:

Let interest rate of $14000 be x%

and Interest rate for $7000 be y %

According to the first condition

14000 * x% - 7000 * y% = 595

multiply by 100

14000x-7000y = 59500

/700

20x-10y=85.................(1)

II condition

x%=y%+0.5%

x=y+0.5

x-y=0.5..................................(2)

solve (1) & (2)

20 x -10 y = 85 .............1

Total value

1 x -1 y = 0.50 .............2

Eliminate y

multiply (1)by 1

Multiply (2) by -10

20.00 x -10.00 y = 85.00

-10.00 x + 10.00 y = -5.00

Add the two equations

10.00 x = 80.00

/ 10.00

x = 8.00

plug value of x in (1)

20.00 x -10.00 y = 85.00

160.00 -10.00 y = 85.00

-10.00 y = 85.00 -160.00

-10.00 y = -75.00

y = 7.50

x= 8.00 Interest rate on $14000

y= 7.50 Interest rate on $7000

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

She types:

After lunch she works:

She types:

Total Ali works= 4+4= 8 hours= 480 min

Total Ali types= 11520+8160= 19680 words

Average typing rate= 19680 words/480 min= 41 word/min

Answer:

114

Step-by-step explanation:

Answer:

144Step-by-step explanation:

Answer:

Step-by-step explanation:

relative speed is calculated by the difference of the one with respect to the other

let car be 70 mph

and lorry  be 50 mph../

relative speed of car with respect to lorry = 70-50 = 20 mph

this is ur answer

The speed of the car relative to the lorry when a car and a lorry travelling-

in the same direction : 20 mph

in the opposite direction: 120 mph

"Relative speed is the speed of one body with respect to another."

If two bodies moving with speed m, n (m > n)

in the same direction, then the relative speed = m - n

in the opposite direction, then the relative speed = m + n

Given: - a car is travelling at 70 mph and a lorry is travelling at 50 mph.

If car and a lorry travelling in the same direction then the relative speed would be,

= 70 - 50

= 20 mph

If the car and a lorry travelling in the opposite direction then the relative speed would be,

= 70 + 50

= 120 mph

Therefore, the speed of the car relative to the lorry is 20 mph if they are travelling in the same direction and 120 mph if the car and a lorry travel in opposite direction.

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Answer:

Step-by-step explanation:

Given that,

[tex]n_1=9,x=520,s^2_x=38\\\\n_2=8,y=540,s^2_y=20[/tex]

a) Under null hypothesis H₀ : there is no difference between the variability of night shift and day shift workers

i.e [tex]H_0:\sigma^2_x=\sigma^2_y=\sigma^2[/tex]

Alternative hypothesis [tex]H_1:\sigma_x^2>\sigma_y^2[/tex]

Level of significance = 5% = 0.05

b) The test statistic

[tex]F=\frac{S^2_x}{S_y^2} \sim F(n_1-1,n_2-1)\\\\=\frac{38}{20}\\\\=1.9[/tex]

Table value of [tex]F_{0.05}(n_1-1,n_2-1)[/tex]

[tex]=F_{0.05}(9-1,8-1))\\\\=F_{0.05}(8,7)\\\\=3.726[/tex]

[tex]\therefore F_{calculated}=1.9<Tab F_{0.05}(8,7)=3.726[/tex]

[tex]H_0[/tex] is accepted at 5% level of significance

Therefore ,there is no difference between the variability of night shift and day shift worker

c) The P-value is 0.206356

The result is not significant at P < 0.05

d) At 95% confidence interval

[tex]\frac{\frac{S^2_x}{S^2_y} }{F_{t-\alpha/2(8,7)} } <\frac{\sigma^2_x}{\sigma^2_y}\frac{\frac{S^2_x}{S^2_y} }{F_{\alpha/2(8,7)} } \\\\\Rightarrow\frac{1.9}{F_{t-(0.05/2)}} <\frac{\sigma^2_x}{\sigma^2_y} <\frac{1.9}{F_{(0.05/2)}} \\\\\Rightarrow\frac{1.9}{F_{0.975}} <\frac{\sigma^2_x}{\sigma^2_y} <\frac{1.9}{F_{(0.025)}} \\\\\Rightarrow\frac{1.9}{F_{3.726}} <\frac{\sigma^2_x}{\sigma^2_y} <\frac{1.9}{F_{(0.286)}}[/tex]

Variance -ratio lies betwee (0.51,6.643)

Conclusion: There is not sufficient evidence to support the claim  that the night shift worker show more variability in their output levels, than the day workers at α =0.05

Answer:

Step-by-step explanation:

Hello!

The claim is that the variability of the output levels of the night-shift is greater than the variability in the output levels of the day workers.

Be

X₁: output level of night shift workers.

n₁= 9

X[bar]₁= 520

S₁= 38

X₂: output level of day shift workers.

n₂= 8

X[bar]₂= 540

S₂= 20

Considering both variables have a normal distribution, the parameters of interest are the population variances.

a)

H₀: σ₁² ≤ σ₂²

H₁: σ₁² > σ₂²

b)

To compare both variances you have to conduct a variance ratio test with statistic:

[tex]F= (\frac{S^2_1}{S^2_2} )*(\frac{Sigma^2_1}{Sigma^2_2} )~~F_{n_1-1;n_2-1}[/tex]

[tex]F_{H_0}= (\frac{1444}{400} )*1=3.61[/tex]

c)

The test is one tailed  to the right, the p-value will have the same direction, i.e. it will be in the right tail of the distribution. The F distribution has degrees of freedom:

n₁ - 1= 9 - 1= 8

n₂ - 1= 8 - 1= 7

P(F₈,₇ ≥ 3.61) = 1 - P(F₈,₇ < 3.61) = 1 - 0.9461= 0.0539

The p-value of this test is 0.0539

d)

The CI for the variance ratio is:

[tex][\frac{S^2_1/S_2^2}{F_{n_1-1;n_2-1;1-\alpha /2}}; \frac{S^2_1/S_2^2}{F_{n_1-1;n_2-1;\alpha /2}}][/tex]

[tex]F_{n_1-1;n_2-1;1-\alpha /2}= F_{8;7;0.975}= 4.90[/tex]

[tex]F_{n_1-1;n_2-1;\alpha /2}= F_{8;7;0.025}= 0.22[/tex]

[tex][\frac{1444/400}{4.90}}; \frac{1444/400}{0.22}}][/tex]

[0.736; 16.409]

Using the level of significance complementary to the confidence level of the interval, you can compare it to the p-value calculated in item c.

p-value: 0.0539

α: 0.05

The p-value is less than the significance level, the decision is to reject the null hypothesis. Using a 5% significance level you can conclude that the variance in the output levels of the night shift workers is greater than the variance in the output levels of the day shift workers.

Answer: B.

x ≈2.5

Step-by-step explanation:

[tex]-\left(u\right)^{-1}-6=-u+10[/tex]

[tex]u=8-\sqrt{65},\:u=8+\sqrt{65}[/tex]

[tex]x=\frac{\ln \left(8+\sqrt{65}\right)}{\ln \left(3\right)}[/tex]

x=2.52...

Answer:

x=2.5

Step-by-step explanation:

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:

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

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!

Answer:

Check below, please

Step-by-step explanation:

Step-by-step explanation:

1.For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.)

When the derivative of a function is equal to zero, then it occurs when we have either a local minimum or a local maximum point. So for our x-coordinates we can say

 [tex]f'(x)=0\: at \:x=2, and\: x=-2[/tex]

2. For which values of x is f '(x) positive?

Whenever we have  

 [tex]f'(x)>0[/tex]

then function is increasing. Since if we could start tracing tangent lines over that graph, those tangent lines would point up.

 [tex]f'(x)>0 \:at [-4,-2) \:and\:(2, \infty)[/tex]

3. For which values of x is f '(x) negative?  

On the other hand, every time the function is decreasing its derivative would be negative. The opposite case of the previous explanation. So

 [tex]f'(x) <0 \: at\: [-2,2][/tex]

4.What do these values mean?

 [tex]f(x) \:is \:increasing\:when\:f'(x) >0\\\\f(x)\:is\:decreasing\:when f'(x)<0[/tex]

5.(b) For which values of x is f ''(x) zero?

In its inflection points, i.e. when the concavity of the curve changes. Since the function was not provided. There's no way to be precise, but roughly

at x=-4 and x=4

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