PLEASE gelp me with this, gelp me please oh please gelp me!

answer is 6

Step-by-step explanation:

3x+11=y

y=29

3x+11=29

3x=29-11

3x=18

x=18÷3

x=6

Answer:6

Step-by-step explanation:

3x+11=29

3x=29-11

3x=18

X=18/3

X=6

Consecutive terms in this sequence are separated by a constant c, so if the 4th term is 15, then the next terms would be

5th: 15 + c

6th: (15 + c) + c = 15 + 2c

7th: (15 + 2c) + c = 15 + 3c

and so on. More generally, since any given number in the sequence depends on the number that came before it, we can write the n-th term in terms of the 4th term,

n-th: 15 + (n - 4) c

Then the 9th term in the sequence is

15 + (9 - 4) c = 35

and solving for c gives

15 + 5c = 35   ==>   5c = 20   ==>   c = 4

Then the 15th term would be

15 + (15 - 4)×4 = 15 + 11×4 = 15 + 44 = 59

Answer:

? = 13.6

Step-by-step explanation:

Let the unknown angle be y

so

tan y= p/b

tan y =8/33

y = tan‐¹(8/33)

y = 13.62699486

y = 13.6

Answer:

682

Step-by-step explanation:

191,919 + 262,626

454545 ÷ 666 = 682.5

Thus meaning 682 bags will have 666 berries and one bag will have 333 berries.

Answer:

????

Step-by-step explanation:

as in y = 4.95 + 3.99 or points? if so just draw a horizontal line at 8.94

Answer:

8.33 hours

Step-by-step explanation:

120-45 = 75

75 ÷ 9 = 8.33

Answer:

5765865876+5737555586=11503421462

Answer:

A)

[tex]k=0[/tex]

B)

[tex]\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}[/tex]

C)

[tex]\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}[/tex]

Step-by-step explanation:

We are given the function:

[tex]\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}[/tex]

A)

Given that h(1) = 20, we want to find k.

h(1) = 20 means that h(x) = 20 when x = 1. Substitute:

[tex]\displaystyle (20) = 20e^{k(1)}[/tex]

Simplify:

[tex]1= e^k[/tex]

Anything raised to zero (except for zero) is one. Therefore:

[tex]k=0[/tex]

B)

Given that h(1) = 40, we want to find 2k + 1.

Likewise, this means that h(x) = 40 when x = 1. Substitute:

[tex]\displaystyle (40) = 20e^{k(1)}[/tex]

Simplify:

[tex]\displaystyle 2 = e^{k}[/tex]

We can take the natural log of both sides:

[tex]\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}[/tex]

By definition, ln(e) = 1. Hence:

[tex]\displaystyle k = \ln 2[/tex]

Therefore:

[tex]2k+1 = 2\ln 2+ 1 \approx 2.3863[/tex]

C)

Given that h(1) = 10, we want to find k - 3.

Again, this meas that h(x) = 10 when x = 1. Substitute:

[tex]\displaystyle (10) = 20e^{k(1)}[/tex]

Simplfy:

[tex]\displaystyle \frac{1}{2} = e^k[/tex]

Take the natural log of both sides:

[tex]\displaystyle \ln \frac{1}{2} = k\ln e[/tex]

Therefore:

[tex]\displaystyle k = \ln \frac{1}{2}[/tex]

Therefore:

[tex]\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931[/tex]

Answer:

Step-by-step explanation:

[tex]we \ add \ \ only \ \ units \ we \ do \ not \ need \ the \ rest \\\\ \bf (3\underline 1 + 13\underline2 + 14\underline3 + 41\underline4 + 51\underline5 +15\underline6 + 6\underline1)= \\\\ 1+2+3+4+5+6+1=2\underline 2 \\\\ Answer: C) \ 2[/tex]

Answer:

6'2

Step-by-step explanation:

Answer:

vertical asymptotes

x=6, x=-6

horizontal asymptotes

y=0

zeros (0,0)

Step-by-step explanation:

f(x) = 6x / ( x^2 - 36)

First factor

f(x) = 6x / ( x-6)(x+6)

Since nothing cancels

The vertical asymptotes are when the denominator goes to zero

x-6 = 0   x+6=0

x=6          x= -6

Since the numerator has a smaller power than the denominator (1 < 2), there is an asymptote at y = 0

To find the zeros, we find where the numerator = 0

6x=0

x=0

[tex]\\ \rm\Rrightarrow y=\dfrac{6x}{x^2-36}[/tex]

The h orizontal asymptote

As x has less degree than x²

Vertical asymptote

Answer:

0.5675 = 56.75% probability that at least 22 of the 44 students selected are non-history majors.

Step-by-step explanation:

The students are chosen without replacement from the sample, which means that the hypergeometric distribution is used to solve this question. We are working also with a sample with more than 10 history majors and 10 non-history majors, which mean that the normal approximation can be used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

[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]

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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.

Approximation:

We have to use the mean and the standard deviation of the hypergeometric distribution, that is:

[tex]\mu = \frac{nk}{N}[/tex]

[tex]\sigma = \sqrt{\frac{nk(N-k)(N-n)}{N^2(N-1)}}[/tex]

In this question:

88 + 88 = 176 students, which means that [tex]N = 176[/tex]

88 non-history majors, which means that [tex]k = 88[/tex]

44 students are selected, which means that [tex]n = 44[/tex]

Mean and standard deviation:

[tex]\mu = \frac{44*88}{176} = 22[/tex]

[tex]\sigma = \sqrt{\frac{44*88*(176-88)*(176-44)}{176^2(175-1)}} = 2.88[/tex]

What is the probability that at least 22 of the 44 students selected are non-history majors?

Using continuity correction, as the hypergeometric distribution is discrete and the normal is continuous, this is [tex]P(X \geq 22 - 0.5) = P(X \geq 21.5)[/tex], which is 1 subtracted by the p-value of Z when X = 21.5. So

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

[tex]Z = \frac{21.5 - 22}{2.88}[/tex]

[tex]Z = -0.17[/tex]

[tex]Z = -0.17[/tex] has a p-value of 0.4325

1 - 0.4325 = 0.5675

0.5675 = 56.75% probability that at least 22 of the 44 students selected are non-history majors.

The final amount is $7,615.27

A = P(1 + r/n)^t

Where,

A = Final amount

P = principal = $7, 200

r = interest rate = 2.5% = 0.025

n = number of periods = 4

t = time = 9 years

A = P(1 + r/n)^t

= 7,200(1 + 0.025/4)^9

= 7,200(1 + 0.00625)^9

= 7,200(1.00625)^9

= 7,200(1.0576769512798)

= 7,615.2740492152

Approximately,

A = $7,615.27

https://brainly.com/question/14003110

Answer:

Cos C = a/h

= 21/35

Step-by-step explanation:

since cos is equal to adjacent angle over hypotenuse angle, so from the question we conclude Cos C = 21/35

Answer: 20.5

Step-by-step explanation:

Cos 43 = X/28

O.73 = x/28

(0.73)(28)=20.5

Answer:

y=-3x+4

Step-by-step explanation:

The y intercept is 4 because the line crosses the y axis at the 4 tic mark

The slope will be -3 because the y decreases by 3 every time the x incerases by 1

y=mx+b

y=-3x+4

Leontief input output model (technology matrix) is an economic model that shows the quantitative relationship and sectorial interdependency in a national economy

The responses with  regards to the question are;

(i) The technology matrix for the economy is presented as follows;

[tex]\mathbf{ A} =\left[\begin{array}{ccc}Agric&&Textile\\0.4&&0.1\\&&\\0.1&&0.2\end{array}\right] \begin{array}{ccc}\mathbf{Per \ Unit}\\Agriculture\\\\Textile\end{array}\right][/tex]

(ii) The required gross production of each industry to meet the desired surplus are;

50 units of agriculture and 250 units of textile

The reason the above values are correct is as follows:

(i) The given parameters are;

The industries in the economy = Agricultural industry and textile industry

Units of agricultural input required per unit of agricultural output = 0.4

Units of textile input required per unit of agricultural output = 0.1

Units of agricultural input required per unit of textile output = 0.1

Units of textile input required per unit of textile output = 0.2

Let X represent agriculture, and let Y represent textile, we have;

[tex]Agric \ for \ agric = \dfrac{0.4 \ units \ of \ agriculture}{1\ unit \ of \ agric \ produced} \times X \ Agric \ produced= 0.4 \cdot X[/tex]

[tex]Agric \ for \ textile = \dfrac{0.1 \ units \ of \ agriculture}{1\ unit \ of \ textile \ produced} \times Y \ textile \ produced= 0.1 \cdot Y[/tex]

We also have;

Textile for agriculture = 0.1·X

Textile for textile = 0.2·Y

Therefore;

X = 0.4·X + 0.1·Y

Y = 0.1·X + 0.2·Y

Therefore;

The technology matrix for the economy is presented as follows;

[tex]\mathbf{Technology \ matrix, A} =\left[\begin{array}{ccc}Agric&&Textile\\0.4&&0.1\\&&\\0.1&&0.2\end{array}\right] \begin{array}{ccc}\mathbf{Per \ Unit}\\Agriculture\\\\Textile\end{array}\right][/tex]

(ii)  Let P represent the production vector, and let d represent the demand vector, we have;

[tex]P = \left[\begin{array}{c}X \\Y\end{array}\right][/tex], [tex]d = \left[\begin{array}{c}5 \\195\end{array}\right][/tex]

P = A·P + d

∴ P - A·P = d

Therefore;

[tex]P = \mathbf{ \dfrac{d}{(I - A)}}[/tex]

Where I = The 2 by 2 identity matrix

We get;

[tex]I - A =\left[\begin{array}{ccc}1&&0\\&&\\0&&1\end{array}\right] - \left[\begin{array}{ccc}0.4&&0.1\\&&\\0.1&&0.2\end{array}\right] = \mathbf{\left[\begin{array}{ccc}0.6&&-0.1\\&&\\-0.1&&0.8\end{array}\right]}[/tex]

With the use of a graphing calculator, we have;

[tex]P =\left[\begin{array}{c}X \\Y\end{array}\right] = \dfrac{\left[\begin{array}{c}5 \\195\end{array}\right]}{\left[\begin{array}{ccc}0.6&&-0.1\\&&\\-0.1&&0.8\end{array}\right]} = \left[\begin{array}{ccc}50\\\\\ 250\end{array}\right][/tex]

The required gross product of agriculture, X = 50 units

The required gross product of textile, Y = 250 units

Learn more about the Leontief input output model here:

https://brainly.com/question/15417573

We have that he technology matrix for this economy and the  the gross production of each industry are

a)    [tex]X= \begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}[/tex]

b)    [tex]\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}50\\250\end{vmatrix}[/tex]          

From the Question we have told that

Each unit of agricultural output requires 0.4 unit of agricultural input

Each unit of agricultural output requires 0.1 unit of textiles input.

Each unit of textiles output requires 0.1 unit of agricultural input

Each unit of textiles output requires 0.2 unit of textiles input.

Generally the technology matrix for this economy is given below

With

X =Agricultural industry Gross output

Y= Textile industry Gross Output

Therefore

[tex]X= \begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}[/tex]

b)

From the Question we are told that

Surpluses of 5 units of agricultural products and 195 units of textiles are desired.

Therefore, we have Desired surplus matrix of

[tex]D= \begin{vmatrix}5\\195\end{vmatrix}[/tex]

Generally the Technology equation is mathematically given as

[tex](I-X)\phi=D[/tex]

Where

X =Agricultural industry Gross output

I=A Unit matrix

\phi=Matrix of gross production

Therefore

[tex]\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}-(\begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}))\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}5\\195\end{vmatrix}[/tex]

[tex]\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}50\\250\end{vmatrix}[/tex]

In conclusion

The technology matrix for this economy and the  the gross production of each industry are

[tex]X= \begin{vmatrix}0.4 & 0.1 \\0.1 & 0.2\end{vmatrix}[/tex]

[tex]\begin{vmatrix}A\\T\end{vmatrix}=\begin{vmatrix}50\\250\end{vmatrix}[/tex]         Respectively

In conclusion

https://brainly.com/question/16863924

We can seperate (3b³) into two different parts, the constant and the variable.

The constant (3) and the variable (b) can both be squared and multiplied to get the correct answer, so:

3² = 9

(b³)² = [tex]b^{6}[/tex]

So, [tex](3b^{3})^{2} = 9b^{6}[/tex]

Answer: I think the answer is A

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

12 floors

Step-by-step explanation:

768 ÷ 64 = 12.

Answer:

12

Step-by-step explanation:

768 divided by 64 =12

Answer:

Step-by-step explanation:

a₁ = 15

a₂/a₁ = 45/15 = 3

a₃/a₂ = 135/45 = 3

...

It is a geometric sequence with a common ratio r=3.

Sum of first 12 terms = a₁·(1-r¹²)/(1-r)

= 15·(1-3¹²)/(1-3)

= 15·(-531,441)/(-2)

= 3,985,800

Answer:

210 miles in 1 hour

Step-by-step explanation:

steps are in picture

Answer:

The number is 6.

Step-by-step explanation:

[tex]4x-13=x+5\\3x-13=5\\3x=18\\x=6[/tex]

Answer:

Step-by-step explanation:

210=2x3x5x7

280=2x2x2x5x7

360=2x2x2x3x3x5

Answer:

210= 2×3×5×7

280=2×2×2×5×7

360=2×2×2×3×3×5

common factors=2×2×2×3×5×7=840

uncommon factors=3

L.C.M=Common factors× uncommon factors

L.C.M=840×3

L.C.M=2520

Step-by-step explanation:

i hope it will be helpful

plzz mark as brainliest

Answer:

A, B, and E.

Step-by-step explanation:

A. 5^x * 5^x

= 5^x+x

=5^(2)(x)

=25^x

B. 5^2x

=5^(2)(x)

=25^x

C. 5*5^2x

=5^1+2x

D. 5*5^x

=5^1+x

E. (5*5)^x

=5^x*5^x

=5^(2)(x)

=25^x

F. 5^2*5^x

=5^2+x

Answer:

The probability tree is;

                            0.95        [tex](+)[/tex]

                   [tex](S)[/tex]

          0.1              0.05        [tex](-)[/tex]

[  P  ]

          0.9             0.15          [tex](+)[/tex]                                                        

                  [tex](S_{no})[/tex]    

                             0.85         [tex](-)[/tex]        

Step-by-step explanation:

Given the data in the question;

10% of the rugby team members use steroids

so Probability of using steroid; P( use steroid ) = 10% = 0.10

Probability of not using steroid; P( no steroid use ) = 1 - 0.10 = 0.90

Since the test show positive for an athlete who uses steroids, 95% of the time.

Probability of using steroids and testing positive = 95% = 0.95

Probability of using steroids and testing Negative = 1 - 0.95 = 0.05

Also from the test, 15% of all steroid-free individuals also test positive.

so

Probability of not using steroids and testing positive = 15% = 0.15

Probability of not using steroids and testing negative = 1 - 0.15 = 0.85

To set up the probability tree, Let;

[tex](S)[/tex] represent steroid use

[tex](S_{no})[/tex] represent no steroid use

[tex](+)[/tex] represent test positive

[tex](-)[/tex] represent test negative

so we have;

                            0.95        [tex](+)[/tex]

                   [tex](S)[/tex]

          0.1              0.05        [tex](-)[/tex]

[  P  ]

          0.9             0.15          [tex](+)[/tex]                                                        

                  [tex](S_{no})[/tex]    

                             0.85         [tex](-)[/tex]