How is the series 6+13+20+...+111 represented in summation notation?

Answer:

68.3 degrees

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan I = opp side / adj side

tan I = sqrt(82) / sqrt(13)

tan I = sqrt(82/13)

Taking the inverse tan of each side

tan ^-1 ( tan I) = tan ^-1( sqrt(82/13))

I = 68.2892

Rounding to the nearest tenth

I = 68.3 degrees

Answer: 58

Step-by-step explanation:

its 58 because chickens have two feet each so divide 2 %  166 and its 58

because each chicken has 2 legs count the 2 legs up to 116 then u get ur answer

Answer:

0.1088 or 10.88%

Step-by-step explanation:

q = 1 - 0.35 = 0.65

P(X=2) = 12C2 × (0.35)² × (0.65)¹⁰

= 0.1088

Answer:

Step-by-step explanation:

a) categorical

b) add all of the numbers and divide by how many numbers there were.

c) outliers means any that were far away from the rest of the data

d) not entirely, you can make an estimate based on it, but nat an exact answer.

Answer:

Step-by-step explanation:

Side=AC=9[tex]\sqrt{2}[/tex]

Side AB= x

Hypotenuse =CB= y

Side AB = 9[tex]\sqrt{2}[/tex]

Hypotenuse CB = 36

Answer:

0.0022 = 0.22% probability that he randomly plants the trees so that all 6 redwoods are next to each other and all 6 pine trees are next to each other.

Step-by-step explanation:

The trees are arranged, so the arrangements formula is used to solve this question. Also, a probability is the number of desired outcomes divided by the number of total outcomes.

Arrangements formula:

The number of possible arrangements of n elements is given by:

[tex]A_n = n![/tex]

Desired outcomes:

Two cases:

6 redwoods(6! ways) then the 6 pine trees(6! ways)

6 pine trees(6! ways) then the 6 redwoods(6! ways)

So

[tex]D = 2*6!*6![/tex]

Total outcomes:

12 trees, so:

[tex]D = 12![/tex]

What is the probability that he randomly plants the trees so that all 6 redwoods are next to each other and all 6 pine trees are next to each other?

[tex]p = \frac{D}{T} = \frac{2*6!*6!}{12!} = 0.0022[/tex]

0.0022 = 0.22% probability that he randomly plants the trees so that all 6 redwoods are next to each other and all 6 pine trees are next to each other.

Answer:

0.0022 = 0.22% probability that she randomly plants the trees so that all 6 pine trees are next to each other and all 6 willows are next to each other.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this question, the elements are arranged, so we have to use the arrangements formula.

Arrangements formula:

The number of possible arrangements of n elements is:

[tex]A_{n} = n![/tex]

Desired outcomes:

Pine trees(6!) then the willows(6!) or

Willows(6!) then the pine trees(6!). So

[tex]D = 2*6!*6! = 1036800 [/tex]

Total outcomes:

12 trees, so:

[tex]T = 12! = 479001600 [/tex]

What is the probability that she randomly plants the trees so that all 6 pine trees are next to each other and all 6 willows are next to each other?

[tex]p = \frac{D}{T} = \frac{1036800 }{479001600 } = 0.0022[/tex]

0.0022 = 0.22% probability that she randomly plants the trees so that all 6 pine trees are next to each other and all 6 willows are next to each other.

9514 1404 393

Answer:

  3×10^8 m/s

Step-by-step explanation:

The desired speed is ...

  [tex]\dfrac{\dfrac{9.45\times10^{15}\text{ m}}{\text{yr}}}{\dfrac{3.15\times10^7\text{ s}}{\text{yr}}}=\dfrac{9.45}{3.15}\times10^{15-7}\text{ m/s}=\boxed{3.00\times10^8\text{ m/s}}[/tex]

__

Your calculator can help you figure this out.

Answer:

Step-by-step explanation:

if it's only true or false there are 2¹⁷=131072 outcomes

if it's true, false, or blank there are 3¹⁷=129140163 outcomes

Volume = πr²h

Radius = 3yd

Height = 12yd

Take π = 22/7

Volume = 22/7×3×3×12

= 2376/7

= 339.4285714yd³

Rounding off to nearest tenth

= 339.43yd³

Answered by Gauthmath must click thanks and mark brainliest

Answer:

Hence the answer is TRUE.

Step-by-step explanation:

If event A is taking 15 or more minutes to urge to figure tomorrow and event B is taking but a quarter-hour to urge to figure tomorrow, then events A and B must be complimentary events. this is often because the occurring of 1 is going to be precisely the opposite of the occurring of the opposite event and that they cannot occur simultaneously. In other words, events A and B are mutually exclusive and exhaustive.  

Mathematically,  

P(A) + P(B) = 1.

Answer:

200

Step-by-step explanation:

Break down 25 = 5*5

Break down 8 = 2*2*2

They have no common factors

The least common multiple is

5*5*2*2*2 = 25*8 = 200

Answer:

200

Step-by-step explanation:

list the factors of 25: 5,5

factors of 8:2,2,2,

Answer:

[tex]\displaystyle\frac{19,683}{200,000}\text{ or }\approx 9.84\%[/tex]

Step-by-step explanation:

For each planted seed, there is a 90% chance that it grows into a healthy plant, which means that there is a [tex]100\%-90\%=10\%[/tex] chance it does not grow into a healthy plant.

Since we are planting 6 seeds, we want to choose 2 that do not grow and 4 that do grow:

[tex]\displaystyle \frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}[/tex]

However, this is only one case that meets the conditions. We can choose any 2 out of the 6 seeds to be the ones that don't grow into a healthy plant, not just the first and second ones. Therefore, we need to multiply this by number of ways we can choose 2 things from 6 (6 choose 2):

[tex]\displaystyle \binom{6}{2}=\frac{6\cdot 5}{2!}=\frac{30}{2}=15[/tex]

Therefore, we have:

[tex]\displaystyle\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \binom{6}{2},\\\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot 15,\\\\P(\text{exactly 2 don't grow})=\boxed{\frac{19,683}{200,000}}\approx 9.84\%[/tex]

Answer:

[tex] {?}^{?} However, this is only one case that meets the conditions. We can choose any 2 out of the 6 seeds to be the ones that don't grow into a healthy plant, not just the first and second ones. Therefore, we need to multiply this by number of ways we can choose 2 things from 6 (6 choose 2):

\displaystyle \binom{6}{2}=\frac{6\cdot 5}{2!}=\frac{30}{2}=15(26)=2!6⋅5=230=15

Therefore, we have:

\begin{gathered}\displaystyle\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \binom{6}{2},\\\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot 15,\\\\P(\text{exactly 2 don't grow})=\boxed{\frac{19,683}{200,000}}\approx 9.84\%\end{gathered}P(exactly 2 don’t grow)=101⋅101⋅109⋅109⋅109⋅109⋅(26),P(exactly 2 don’t grow)=101⋅101⋅109⋅109⋅109⋅109⋅15,P(exactly 2 don’t grow)=200,00019,683≈9.84%

[/tex]

Answer:

50%

Step-by-step explanation:

35 is halve of 70 therefore it is 50%

hope it helps u...........

Answer:

a) 0.0081 = 0.81% probability that he receives no job offer

b) He expects to get 2.8 job offers.

c) 0.0837 = 8.37% probability that more than half of the firms he applied do not make him any offer.

d) Each job must be independent of other jobs. Additionaly, if [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the normal approximation to the binomial distribution can be used.

e) 0.2401 = 24.01% probability of him securing more than 3 offers.

Step-by-step explanation:

For each application, there are only two possible outcomes. Either he gets an offer, or he does not. The probability of getting an offer for a job is independent of any other job, which means that the binomial probability distribution is used 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.

He can expect to receive a job offer from 70% of the firms to which he applies.

This means that [tex]p = 0.7[/tex]

The student decides to apply to only four firms.

This means that [tex]n = 4[/tex]

(a) What is the probability that he receives no job offer?

This is [tex]P(X = 0)[/tex]. So

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

[tex]P(X = 0) = C_{4,0}.(0.7)^{0}.(0.3)^{4} = 0.0081[/tex]

0.0081 = 0.81% probability that he receives no job offer.

(b) How many job offers he expects to get?

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

In this question:

[tex]E(X) = 4(0.7) = 2.8[/tex]

He expects to get 2.8 job offers.

(c) What is the probability that more than half of the firms he applied do not make him any offer?

Less than 2 offers, which is:

[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]

So

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

[tex]P(X = 0) = C_{4,0}.(0.7)^{0}.(0.3)^{4} = 0.0081[/tex]

[tex]P(X = 1) = C_{4,1}.(0.7)^{1}.(0.3)^{3} = 0.0756[/tex]

Then

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0081 + 0.0756 = 0.0837[/tex]

0.0837 = 8.37% probability that more than half of the firms he applied do not make him any offer.

(d) What assumptions do you need to make to find the probabilities? To increase the chance of securing more job offers, the student decides to apply to as many companies as possible, he sent out 60 applications to all different accounting firms.

Each job must be independent of other jobs. Additionaly, if [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the normal approximation to the binomial distribution can be used.

(e) What is the probability of him securing more than 3 offers?

Between 4 and n, since n is 4, 4 offers, so:

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

[tex]P(X = 4) = C_{4,4}.(0.7)^{4}.(0.3)^{0} = 0.2401[/tex]

0.2401 = 24.01% probability of him securing more than 3 offers.

Answer:

f(1) = -2

f(x) =1 when x=0 or x=-2

Step-by-step explanation:

f(1) is the y value when x=1

f(1) = -2

f(x) = 1 means find the x value when y=1

when y =1, x =0 and -2

Answer:

5880 ways

Step-by-step explanation:

For selections like this, we solve using the combination theory. Recall that

nCr = n!/(n-r)!r!

Hence given to find the number of ways 6 gentleman and 4 ladies can be choosen out of 10 gentleman and 8 ladies,

= 10C6 * 8C4

= 10!/(10-6)!6! * 8!/(8-6)!6!

= 10 * 9 * 8 * 7 * 6!/4 *3 *2 * 6! * 8 * 7 * 6!/2 * 6!

= 210 * 28

= 5880 ways

The arrangement can be done in 5880 ways

========================================================

Explanation:

At 95% confidence, the z critical value is roughly z = 1.960

The population standard deviation is given to be sigma = 3.7

The error is E = 2 since we want to be within 2 inches of the population mean mu

The min sample size needed is:

n = (z*sigma/E)^2

n = (1.960*3.7/2)^2

n = 13.147876

n = 14

We always round up to the nearest whole number to ensure that we clear the hurdle (otherwise, the sample is too small). It doesn't matter that we're closer to 13 than to 14.

Answer:

108.82

Step-by-step explanation:

Answer:

I don't see the problem.

Step-by-step explanation:

Answer:

I hope it help and I guess it is correct

It looks like the integral you want to find is

[tex]\displaystyle \int_C x^2y\,\mathrm dx - xy^2\,\mathrm dy[/tex]

where C is the circle x ² + y ² = 4. By Green's theorem, the line integral is equivalent to a double integral over the disk x ² + y ² ≤ 4, namely

[tex]\displaystyle \iint\limits_{x^2+y^2\le4}\frac{\partial(-xy^2)}{\partial x}-\frac{\partial(x^2y)}{\partial y}\,\mathrm dx\,\mathrm dy = -\iint\limits_{x^2+y^2\le4}(x^2+y^2)\,\mathrm dx\,\mathrm dy[/tex]

To compute the remaining integral, convert to polar coordinates. We take

x = r cos(t )

y = r sin(t )

x ² + y ² = r ²

dx dy = r dr dt

Then

[tex]\displaystyle \int_C x^2y\,\mathrm dx - xy^2\,\mathrm dy = -\int_0^{2\pi}\int_0^2 r^3\,\mathrm dr\,\mathrm dt \\\\ = -2\pi\int_0^2 r^3\,\mathrm dr \\\\ = -\frac\pi2 r^4\bigg|_{r=0}^{r=2} \\\\ = \boxed{-8\pi}[/tex]

explainion:

[tex] {a}^{2} + {b}^{2} = {c}^{2} [/tex]

Use The (Pythagorean Theorem) to find the length of any side of a right triangle. Form it like its shown in picture above. Follow the instructions that also shown in the picture above.

Answer:

[tex]43.13 = 5.25h + 9[/tex]

Step-by-step explanation:

Let's solve this by making an equation.

$9 for the helmet, and $5.25 per hour.

h will stand for hours, C will stand for Amanda's cost.

[tex]C = 5.25h + 9[/tex]

Now, substitute in what we learned from the problem.

[tex]43.13 = 5.25h + 9[/tex]

This is an equation you can use to solve for the hours.

Given:

The given expression is:

[tex]6x^2y-3xy-24xy^2+12y^2[/tex]

To find:

Part A: The expression by factoring out the greatest common factor.

Part B: Factor the entire expression completely.

Solution:

Part A:

We have,

[tex]6x^2y-3xy-24xy^2+12y^2[/tex]

Taking out the highest common factor 3y, we get

[tex]=3y(2x^2-x-8xy+4y)[/tex]

Therefore, the required expression is [tex]3y(2x^2-x-8xy+4y)[/tex].

Part B:

From part A, we have,

[tex]3y(2x^2-x-8xy+4y)[/tex]

By grouping method, we get

[tex]=3y(x(2x-1)-4y(2x-1))[/tex]

[tex]=3y(x-4y)(2x-1)[/tex]

Therefore, the required factored form of the given expression is [tex]3y(x-4y)(2x-1)[/tex].

Answer:

distributive

Step-by-step explanation:

a(b + c)=ab + ac

it's distributive one

Multiplying block matrices works just like multiplication between regular matrices, provided that component matrices have the right sizes.

[tex]\begin{bmatrix}\mathbf I&\mathbf 0\\\mathbf K&\mathbf I\end{bmatrix}\begin{bmatrix}\mathbf W&\mathbf X\\\mathbf Y&\mathbf Z\end{bmatrix} = \begin{bmatrix}\mathbf{IW}+\mathbf{0Y}&\mathbf{IX}+\mathbf{0Z}\\\mathbf{KW}+\mathbf{IY}&\mathbf{KX}+\mathbf{IZ}\end{bmatrix}[/tex]

[tex]\begin{bmatrix}\mathbf I&\mathbf 0\\\mathbf K&\mathbf I\end{bmatrix}\begin{bmatrix}\mathbf W&\mathbf X\\\mathbf Y&\mathbf Z\end{bmatrix} = \begin{bmatrix}\mathbf W+\mathbf 0&\mathbf X+\mathbf 0\\\mathbf{KW}+\mathbf Y&\mathbf{KX}+\mathbf Z\end{bmatrix}[/tex]

[tex]\begin{bmatrix}\mathbf I&\mathbf 0\\\mathbf K&\mathbf I\end{bmatrix}\begin{bmatrix}\mathbf W&\mathbf X\\\mathbf Y&\mathbf Z\end{bmatrix} = \begin{bmatrix}\mathbf W&\mathbf X\\\mathbf{KW}+\mathbf Y&\mathbf{KX}+\mathbf Z\end{bmatrix}[/tex]

(I assume I is the identity matrix and 0 is the zero matrix.)