help me with this please pleaseee!

Answer:

Step-by-step explanation:

Combination of 3 teachers out of 12:

Combination of 4 students out of 43:

Total combinations:

Step-by-step explanation:

2tan 30° / 1 + tan² 30° =

2(⅓√3) /1 + (⅓√3)² =

⅔√3 / 1+ ⅓ =

⅔√3 / 4/3 =

2/4 √3 =

½√3 = sin 60° (proven)

Answer:

9

Step-by-step explanation:

Answer:

1.

a. 2

b. 0.1353 probability that exactly 0 customers will arrive during a five-minute period, 0.2707 that exactly 1 customer will arrive, 0.2707 that exactly 2 customers will arrive and 0.1805 that exactly 3 customers will arrive.

c. 0.1428 = 14.28% probability that delays will occur.

2.

a. 0.4512 = 45.12% probability that the service time is one minute or less.

b. 0.6988 = 69.88% probability that the service time is two minutes or less.

c. 0.3012 = 30.12% probability that the service time is more than two minutes.

Step-by-step explanation:

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

[tex]f(x) = \mu e^{-\mu x}[/tex]

In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.

The probability that x is lower or equal to a is given by:

[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]

Which has the following solution:

[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]

The probability of finding a value higher than x is:

[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]

Question 1:

a. What is the mean or expected number of customers that will arrive in a five-minute period?

0.4 customers per minute, so for 5 minutes:

[tex]\mu = 0.4*5 = 2[/tex]

So 2 is the answer.

Question b:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353[/tex]

[tex]P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707[/tex]

[tex]P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707[/tex]

[tex]P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1805[/tex]

0.1353 probability that exactly 0 customers will arrive during a five-minute period, 0.2707 that exactly 1 customer will arrive, 0.2707 that exactly 2 customers will arrive and 0.1805 that exactly 3 customers will arrive.

Question c:

This is:

[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]

In which:

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

The values we have in item b, so:

[tex]P(X \leq 3) = 0.1353 + 0.2707 + 0.2707 + 0.1805 = 0.8572[/tex]

[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.8572 = 0.1428[/tex]

0.1428 = 14.28% probability that delays will occur.

Question 2:

[tex]\mu = 0.6[/tex]

a. What is the probability that the service time is one minute or less?

[tex]P(X \leq 1) = 1 - e^{-0.6} = 0.4512[/tex]

0.4512 = 45.12% probability that the service time is one minute or less.

b. What is the probability that the service time is two minutes or less?

[tex]P(X \leq 2) = 1 - e^{-0.6(2)} = 1 - e^{-1.2} = 0.6988[/tex]

0.6988 = 69.88% probability that the service time is two minutes or less.

c. What is the probability that the service time is more than two minutes?

[tex]P(X > 2) = e^{-1.2} = 0.3012[/tex]

0.3012 = 30.12% probability that the service time is more than two minutes.

Answer:

24

Step-by-step explanation:

divide the area in 2 regions

4 x 2 = 8 (area of one region)

4 x 4 = 16 (area of second region)

8 + 16 = 24 (sum of areas of the two regions)

Answer:

D

Step-by-step explanation:

[tex]y - \frac{3}{3} + 12[/tex]

[tex]y - 1 + 12[/tex]

[tex]y + 11[/tex]

Answer:

The right solution is "0.5".

Step-by-step explanation:

According to the question,

P(pay with the sponsored credit card | order exceeds $110)

= [tex]\frac{P(Pay \ with \ the \ sponsored \ credit\ card\ and\ order\ exceeds\ 110)}{P(order \ exceeds \ 110)}[/tex]

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

By putting the values, we get

= [tex]\frac{0.17}{0.34}[/tex]

= [tex]0.5[/tex]

Thus, the above is the right solution.

Answer:

wheres the underline pls let me know what is underlined ill answer it on comment

The correct statement about the flow of information from the adjusted trial balance on the end-of-period spreadsheet is A. The revenue and expense account balances flow into the income statement.

This refers to the general ledger balance after some changes have been done an account balance such as accrued expenses, depreciation, etc.

Therefore, we can see that from the complete information, the statement that is false about the adjusted trial balance on the end-of-period spreadsheet is option A because the revenue and expense account balances does not flow into the income statement.

The other options from the complete text are:

Read more about adjusted trial balance here:

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Answer:

huh ano yan huhu paki ayos ng sagot

Step-by-step explanation:

hahahhaa

Answer:

c. 1.29

Step-by-step explanation:

Z-score:

In a set 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]

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 p-value, we get the probability that the value of the measure is greater than X.

Andy 70578 60000 8200

This means that [tex]X = 70578, \mu = 60000, \sigma = 8200[/tex]

What is the z-score of Andy's salary?

This is Z, so:

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

[tex]Z = \frac{70578 - 60000}{8200}[/tex]

[tex]Z = 1.29[/tex]

So the correct answer is given by option c.

Step-by-step explanation:

here is the answer to your question

Answer:

[tex] {f}^{ - 1} (x) = \frac{x}{3} + \frac{7}{3} [/tex]

hence option d is the correct option.

Answer:

Option C is answer.

Step-by-step explanation:

Hey there!

Given;

f(x) = (3-x) /7

Let f(x) be "y".

y = (3-x) /7

Interchanging "x" and "y".

x = (3-y)/7

7x = 3-y

y = 3-7x

Therefore, f'(x) = 3-7x.

Hope it helps!

Answer:

the first wheel ever built

Answer:

The first wheel ever built

Step-by-step explanation

(*) Sorry for my late answer but I hope this helps others that are looking for this.

100% in the test :)

Answer:1

Step-by-step explanation:2

2

The value of a when the function f(x) = 3|xl is written in the standard form of an absolute value function is 3.

An absolute function is defined as a function which consists of an algebraic expression that is within absolute value symbols.

Here,

The standard form of the absolute value function is written by,

f(x) = a|x|

Given that,

f(x) = 3|x|

Comparing this with the standard form, we get,

a|x| = 3|x|

Therefore, a = 3

Hence,

The value of a when the function f(x) = 3|xl is written in the standard form of an absolute value function is 3.

To learn more about absolute function, click:

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Answer:

1326

Step-by-step explanation:

[tex]{52\choose2}=\frac{52!}{(52-2)!2!}=\frac{52!}{50!*2!}=1326[/tex]

Answer: 10

Step-by-step explanation:

4 x 2 x 5 = 40

9 + 2 + 7 + 8 + 4 = 30

40 - 30 = 10

= 10

9514 1404 393

Answer:

  h = -4.79(t -0.58)^2 +3.26

Step-by-step explanation:

The coordinates (t1, h1) are the time and height at the maximum. Then 'a' can be found from ...

  h = a(t -t1)^2 +h1

  1.65 = a(0 -0.58)^2 +3.26

  -1.61 = 0.3364a . . . . . subtract 3.26

  -4.786 = a . . . . . . . divide by the coefficient of a

The equation is ...

  h = -4.79(t -0.58)^2 +3.26

Let y = √x, so that y ² = x and 2y dy = dx. Then the integral becomes

[tex]\displaystyle \int_{25}^{81} \frac{\sqrt x}{x-1}\,\mathrm dx = \int_{\sqrt{25}}^{\sqrt{81}} \frac y{y^2-1}(2y\,\mathrm dy) = 2 \int_5^9 \frac{y^2}{y^2-1}\,\mathrm dy[/tex]

Now,

y ² / (y ² - 1) = 1 + 1 / (y ² - 1) = 1 + 1/2 (1/(y - 1) - 1/(y + 1))

so integrating gives us

[tex]\displaystyle 2\int_5^9\frac{y^2}{y^2-1}\,\mathrm dy= \int_5^9\left(2+\frac1{y-1}-\frac1{y+1}\right)\,\mathrm dy \\\\= (2y+\ln|y-1|-\ln|y+1|)\bigg|_5^9 \\\\= \boxed{8+\ln\left(\dfrac65\right)}[/tex]

Answer:

5.5inches

Step-by-step explanation:

1331/8=166.375

then length of a side is = cubic root of 166.375

=³√166.375

5.5

Answer:

13,125 ducks

Step-by-step explanation:

The ratio of ducks:geese on the first day was:

175:63

On the last day (end of migration), he counted 4,725 geese.

To find the number of ducks using the same ratio, we are first going to divide 4,725 by 63 to find what number all the ducks and geese multiplied by:

4,725/63 = 75

The geese multiplied by 75. This means the ducks also multiplied by 75:

175*75 = 13,125

Barnaby counted 13,125 ducks.

Hope it helps (●'◡'●)

Answer:

107%

Step-by-step explanation:

Round up all possible algorithms

Answer:

[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]

[tex]P(x < 70) = 0.225[/tex]

Step-by-step explanation:

Given

[tex]Lower = 30[/tex]

[tex]Width = 10[/tex]

Solving (a): The relative frequency table

First, we construct the frequency table using the given parameters.

[tex]\begin{array}{cc}{Class} & {Frequency} &{30-39} & {1} & {40-49} & {1} & {50 - 59} & {2} & {60 - 69} & {5} & {70 - 79} & {13} & {80 - 89} & {10} & {90 - 99} & {8} & {Total} & {40}\ \end{array}[/tex]

The relative frequency (RF) is calculated as:

[tex]RF = \frac{Frequency}{Total}[/tex]

Using the above formula to calculate the relative frequency, the relative frequency table is:

[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]

Solving (b): [tex]P(x < 70)[/tex]

To do this, we add up the relative frequencies of classes less than 70.

i.e.

[tex]P(x < 70) = [30 - 39] + [40 - 49] + [50 - 59] + [60 - 69][/tex]

So, we have:

[tex]P(x < 70) = 0.025 + 0.025 + 0.050 + 0.125[/tex]

[tex]P(x < 70) = 0.225[/tex]

Answer:

23

Step-by-step explanation:

Answer:

The function represents the sequence is - 3 n + 23.

Step-by-step explanation:

20. 17, 14, 11, 8,......

Here, the first term is

a = 20

Common difference, d = -3

Let the nth term is Tn.

Tn = a + (n -1) d

Tn = 20 + (n -1) x (-3)

Tn = 20 - 3 n + 3

Tn = 23 - 3 n = - 3 n + 23

So, the function represents the sequence is - 3 n + 23.

Answer:

Y'all know what it is already, but I want points, so: f(n) = -3n + 23

Answer:

0，0

Step-by-step explanation:

n=2，

[tex] {2}^{2} - 2 - 2 = 0 [/tex]

[tex] {2}^{2} - 5 \times 2 + 6 = 0[/tex]

Answer:

3

Step-by-step explanation:

make a column of x ,f, fx

then write income in x and no.of workers in f

andthen multiply both just like 100*3 ,100*2, 300*p, 400*2,500*1 write its answer fx

add the all fx and use this formula

mean =fx /n

260=adding total of fx divide by 5

Repeat same formula in no 2

Answer:

This is a multiplication of functions g and f, and these functions have no restrictions(such as a even root or a fraction), and thus [tex](g \mult f)(0) = g(0)f(0) = -4(1) = -4[/tex]

Step-by-step explanation:

We are given the following functions:

[tex]f(x) = 1[/tex]

[tex]g(x) = x - 4[/tex]

Can you evaluate (g•f)(0)?

This is a multiplication of functions g and f, and these functions have no restrictions(such as a even root or a fraction), and thus [tex](g \mult f)(0) = g(0)f(0) = -4(1) = -4[/tex]

Answer:

To evaluate the composition, you need to find the value of function f first. But, f(0) is 1 over 0, and division by 0 is undefined. Therefore, you cannot find the value of the composition.

You must evaluate the function f first.

Division by 0 is undefined.

The composition cannot be evaluated.

Answer:

radius of 80cm is the answer