Pls help!!

Find the interquartile range.

Answer:

Step-by-step explanation:

From the given information:

The total number of wine = 9 + 10 + 12 = 31

(1)

The number of distinct sequences used for serving any five wines can be estimated by using the permutation of the number of total wines with the number of wines served.

i.e

= [tex]^{31}P_5[/tex]

[tex]=\dfrac{31!}{(31-5)!}[/tex]

[tex]=\dfrac{31!}{(26)!}[/tex]

[tex]=\dfrac{31\times 30\times 29\times 28\times 27\times 26!}{(26)!}[/tex]

= 20389320

(2)

If the first two wines served = zinfandel and the last three is either merlot or cabernet;

Then, the no of ways we can achieve this is:

= [tex]^9P_2\times ^{22}P_3[/tex]

[tex]= \dfrac{9!}{(9-2)!}\times \dfrac{22!}{(22-3)!}[/tex]

[tex]= \dfrac{9!}{(7)!}\times \dfrac{22!}{(19)!}[/tex]

[tex]= \dfrac{9*8*7!}{(7)!}\times \dfrac{22*21*20*19!}{(19)!}[/tex]

= 665280

(3)

The probability that no zinfandel is served is computed as follows:

Total wines (with zinfandel exclusion) = 31 - 9 = 22

Now;

the required probability is:

[tex]= \dfrac{^{22}P_5 }{^{31}P_5}[/tex]

[tex]= \dfrac{\dfrac{22!}{(22-5)!} } {\dfrac{31!}{(31-5)!} }[/tex]

[tex]= \dfrac{\dfrac{22!}{17!} } {\dfrac{31!}{(26)! }}[/tex]

[tex]= \dfrac{(\dfrac{22*21*20*19*18*17!}{17!})} {(\dfrac{31*30*29*28*27*26!}{(26)! })}[/tex]

= 0.1549

≅ 0.155

Answer:

The right answer is "1.818".

Step-by-step explanation:

The given values are:

a = 0.1

[tex]\bar{x} = 94[/tex]

[tex]\mu = 90[/tex]

Now,

The test statistic will be:

⇒  [tex]t=\frac{\bar{x}-\mu}{\frac{s}{vn} }[/tex]

On substituting the given values, we get

⇒     [tex]=\frac{94-90}{\frac{22}{10} }[/tex]

⇒     [tex]=\frac{4}{2.2}[/tex]

⇒     [tex]=1.818[/tex]

Answer:

31,680 feet

Step-by-step explanation:

6 * 5280 = 31,680

Answer:

angle classification is acute

side classification is equilateral

Step-by-step explanation:

Answer:

The question and choices are cut off, could you repost it?

Step-by-step explanation:

Answer:

[tex]\angle CAD = 20^\circ[/tex]

Step-by-step explanation:

Given

[tex]\angle CBD = 20^\circ[/tex]

Required

Determine the measure of [tex]\angle CAD[/tex]

To calculate CAD, we have:

[tex]\angle CAD = \angle CBD[/tex] --- angle in the same segment

Hence:

[tex]\angle CAD = 20^\circ[/tex]

Answer:

75

Step-by-step explanation:

Because the number that occurred the most is 75.

Answer:

20

Step-by-step explanation:

5 to 9 = 11

0 to 4 = 9

11 + 9 =  20.

hope this helps

Answer:

Function that fits the points [tex](0,5),\,(2,-13)[/tex] is given by [tex]x+9y=45[/tex]

Step-by-step explanation:

Let [tex](x_1,y_1)=(0,5),\,(x_2,y_2)=(2,-13)[/tex]

Slope of a line joining points [tex](x_1,y_1),\,(x_2,y_2)[/tex] is given by [tex]m=\frac{x_2-x_1}{y_2-y_1}[/tex]

[tex]=\frac{2-0}{-13-5}[/tex]

[tex]=\frac{2}{-18}[/tex]

[tex]=\frac{-1}{9}[/tex]

Equation of a line joining points [tex](x_1,y_1),\,(x_2,y_2)[/tex] is given as follows:

[tex]y-y_1=m (x-x_1)[/tex]

[tex]y-5=\frac{-1}{9}(x-0)\\y-5=\frac{-1}{9}x\\9(y-5)=-x\\9y-45=-x\\x+9y=45[/tex]

Answer:

8

Step-by-step explanation:

Answer:

C

119 + 36 + 25 = 180

Step-by-step explanation:

Answer:

it is proportional

Step-by-step explanation:

Answer:

it is proportional

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

sorry if wrong plz give brainliest thank and like!

Answer:

look at the picture i sent

Step-by-step explanation:

(x²-x+5x-5)(x+2)

(x²+4x-5)(x+2)

x³+2x²+4x²+8x-5x-10

x³+6x²+3x-10

Answer:

a. 0.5 = 50% probability that the driver is selected as a medium-risk driver.

b. 0.3 = 30% probability that he has been classified as high-risk

c. 0.51 = 51% probability that at least one of them has been classified as high-risk.

Step-by-step explanation:

To solve this question, we need to understand conditional probability, for items a and b, and the binomial distribution, for item c.

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

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. If a driver had an accident during the year, find the probability that the driver is selected as a medium-risk driver.

Event A: Had an accident

Event B: Medium-risk driver

Probability of having an accident:

0.01 of 0.6(low risk)

0.05 of 0.3(medium risk)

0.09 of 0.1(high risk)

So

[tex]P(A) = 0.01*0.6 + 0.05*0.3 + 0.09*0.1 = 0.03[/tex]

Probability of having an accident and being a medium risk driver:

0.05 of 0.3. So

[tex]P(A \cap B) = 0.05*0.3 = 0.015[/tex]

Desired probability:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.015}{0.03} = 0.5[/tex]

0.5 = 50% probability that the driver is selected as a medium-risk driver.

b. If a driver who had an accident during the I-year period is selected, what is the probability that he has been classified as high-risk?

Event A: Had an accident

Event B: High risk driver.

From the previous item, we already know that P(A) = 0.03.

Probability of having an accident and being a high risk driver is 0.09 of 0.1. So

[tex]P(A \cap B) = 0.1*0.09 = 0.009[/tex]

The probability is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.009}{0.03} = 0.3[/tex]

0.3 = 30% probability that he has been classified as high-risk

c. If two drivers who had an accident during the I -year period are selected, what is the probability that at least one of them has been classified as high-risk?

0.3 are classified as high risk, which means that [tex]p = 0.3[/tex]

Two accidents mean that [tex]n = 2[/tex]

This probability is:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which

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

[tex]P(X = 0) = C_{2,0}.(0.3)^{0}.(0.7)^{2} = 0.49[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.49 = 0.51[/tex]

0.51 = 51% probability that at least one of them has been classified as high-risk.

Answer: The order is

29,  36, 42, 49, 52, 59

Step-by-step explanation: I hope this is what you were looking for let me know if you need a better answer

Mean- 42.2

Median- 45.5

Mode- 29

Midrange- ??

Range- 29

Minimum- 29

Maximum- 57

Sum- 422

Hope this helped! have a great day! :D

Answer:

A = 46%

B = 72%

Step-by-step explanation:

This question is incomplete, the complete question is;

The times when goals are scored in hockey are modeled as a Poisson process in Morrison (1976). For such a process, assume that the average time between goals is 15 minutes.

(The parameter of the hockey Poisson Process is lambda = 1/15 )

(i) In a 60-minute game, find the probability that a fourth goal occurs in the last 5 minutes of the game?

(ii) Assume that at least three goals are scored in a game. What is the mean time of the third goal?

Answer:

i) the probability that a fourth goal occurs in the last 5 minutes of the game is 0.068

ii) The mean time of the third goal is 33.5 minutes

Step-by-step explanation:

Given the data in the question;

The parameter of the hockey Poisson Process λ = 1/15

i)

Let us represent the probability of a fourth goal in the last 5 min in a 60 min game with X.

Thus, we find the probability that X is greater than ( 60min - 5min)  and less than or equal to 60min

so;

p( 55 < X ≤ 60 ) = [tex]\frac{1}{6} \int\limits^{60}_{55} (\frac{1}{15})^4 ( t^3) (e^{-t}) dt[/tex]

p( 55 < X ≤ 60 ) = 0.06766 ≈ 0.068

Therefore, the probability that a fourth goal occurs in the last 5 minutes of the game is 0.068

ii)

Also let us represent the probability that at least 3 goals are scored in the game with X

Now, the mean time of the third goal will be;

P(X|X < 60 ) = [tex]\frac{1}{P(X<60)} \int\limits^0_{60} t\frac{(1/15)^3 (t^2) ( e^{-t/15}) }{2} dt[/tex]

P(X|X < 60 ) = 25.49 / 0.76

P(X|X < 60 ) = 33.539 ≈ 33.5

Therefore, the mean time of the third goal is 33.5 minutes

Answer:

A≈615.75m²

Step-by-step explanation:

Solution

A=4πr2=4·π·72≈615.75216m²

Hope this helped!!!

count the number of triangles,calculate the area then multiply by the no of triangles and again back to rectangle

Answer:

The sales tax would be $6.71 :)

Step-by-step explanation:

110 x 0.061 = 6.71

116.71

Step-by-step explanation:

if the tax rate is 6.1 that wiild equal 6.71 then you would add 6.71 to 110 which the answer would be 116.71

Answer:

14.7 cm

Step-by-step explanation:

Using Pythagoras Theorem,

The length of the other leg

[tex] = \sqrt{ {19}^{2} - {12}^{2} } [/tex]

[tex] = \sqrt{217} [/tex]

[tex] = 14.730919...[/tex]

= 14.7 cm (rounded to the nearest tenth)

60% of 45 would be 27

He pays his parents $27 each week

Divide 162 and 27

162/27=6

It will take 6 weeks for him to pay back his parents

Answer:

5.83

Step-by-step explanation:

Using Pythagoras Theorem,

[tex] \sqrt{ {5}^{2} + {3}^{2} } [/tex]

[tex] = \sqrt{34} [/tex]

[tex] = 5.83095...[/tex]

= 5.83 units (rounded to the nearest hundredth) +

Answer:

I think No..............

Step-by-step explanation:

they are not right angle .

hope it helps.

Answer:

[tex]length = (8 + x) \: feet\\ width = x \: feet \\ area = length \times width \\ 33 = (8 + x)x \\ 33 = 8x + {x}^{2} \\ {x}^{2} + 8x - 33 = 0 \\ (x - 3)(x + 11) = 0 \\ (x + 11) \: is \: neglected \: since \: it \: is \: negative \\ therefore : x = 3 \\ dimensions \\ length = (8 + 3) = 11 \: feet \\ width = 3 \: feet[/tex]

Answer:

whats the answer? And equation?

Step-by-step explanation: