I really need help big time thank you

60 mL=55% alcohol

?mL=85% alcohol

85×60÷55

=92.727 mL

I think do like this....I'm not sure

Hope this help you

9514 1404 393

Answer:

  120 mL

Step-by-step explanation:

Let x represent the amount of 85% solution needed. Then the total amount of alcohol in the mix is ...

  0.55(60) +0.85(x) = 0.75(60 +x)

  33 + 0.85x = 45 +0.75x . . . . . . . eliminate parentheses

  0.85x = 12 +0.75x . . . . . . . . . .subtract 33

  0.10x = 12 . . . . . . . . . . . . . . subtract 0.75x

  x = 120 . . . . . . . . . . . . . . multiply by 10

You will need 120 mL of the 85% solution.

Answer:

yes, the volume = 16 ft^3

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:

76 natural number not including the first

Hope this helps <3 Comment if you want more thanks and be sure to give brainliest (4 left) <3

There are 78 natural numbers between 68 and 145.

The Natural numbers are defined as it used for counting and are a component of the number system, which includes all positive integers from 1 to infinity. It does not include zero (0).

The set of natural numbers includes only the positive integers, i.e., 1, 2, 3, 4, 5, 6, ……….∞.

Sum of natural numbers are n(n+1)/2

Sn = n(n+1)/2

Hence, this is the formula to calculate sum of 'n' natural numbers.

Given that numbers are 68 and 145

We will start counting from 68 to 144

Number of natural numbers between 68 and 145 = (144 - 67)+1

Number of natural numbers between 68 and 145= 77 +1

Number of natural numbers between 68 and 145 = 78

Hence, there are 78 natural numbers between 68 and 145.

Learn more about Natural numbers here:

https://brainly.com/question/17429689

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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

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

Answer:

$2 for 1 ticket, 7 tickets for $14, $18 for 9 tickets, 10 tickets for $20

Step-by-step explanation:

Since we know that it  $8 for 4 tickets, we can simplify the ratio down to 2:1, meaning that each ticket is $2.

You're welcome and good luck with your classes young one

Answer:

Hello,

Answer : 1400 and 1575

Step-by-step explanation:

Let's say a and b the ywo numbers

[tex]HCF(a,b)=a\vee b=175=5^2*7\\LCM(a,b)=a\wedge b=12600\\\\a*b=(a\vee b)*(a\wedge b)=(2^3*3^2*5^2*7)*(5^2*7)=2^3*3^2*(5^2*7^2)^2\\\\Both\ numbers\ are\ greater\ than\ their HCF\\a=175*k_1\\b=175*k_2\\\\k_1=2^3\ and\ k_2=3^2\\\\a=175*2^3=1400\\b=175*3^2=1575\\\\[/tex]

Answer:

Step-by-step explanation:

Answer:

5.4

Step-by-step explanation:

cos(theta) = B/H

cos(75)=x/21, x=5.4

Answer:

6,78

Step-by-step explanat

ion:data size :10

Sample mean:15

Standard sample deviation :6,782

Answer:

6,78

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

The 95% confidence interval of the population mean, in years, is (4.3, 5.3). 4 years is not part of the confidence interval, which means that it contradicts the fact that 39% of students get their college degree in four years.

-----------------------------

To solve this question, we need to find the confidence interval for the amount of time it takes the students to get the degree.

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

-----------------------------

The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So

df = 80 - 1 = 79

-----------------------------

95% confidence interval

Standard level of confidence, we have to find a value of T, which is found looking at the t table, with 79 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 1.9905.

-----------------------------

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 1.9905\frac{2.2}{\sqrt{80}} = 0.5[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

-----------------------------

The lower end of the interval is the sample mean subtracted by M. So it is 4.8 - 0.3 = 4.3 years.

The upper end of the interval is the sample mean added to M. So it is 4.8 + 0.3 = 5.3 years.

-----------------------------

The 95% confidence interval of the population mean, in years, is (4.3, 5.3). 4 years is not part of the confidence interval, which means that it contradicts the fact that 39% of students get their college degree in four years.

A similar question is given at https://brainly.com/question/24278748

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:

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:

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

Answer:

In a residual plot against x that does NOT suggest we should challenge the assumptions of our regression model, we would expect to see a _____.

c. horizontal band of points centered near 0

Step-by-step explanation:

This residual graph or plot shows the residual values (or the difference between the observed y-value (from scatter plot) and the predicted y-value (from regression equation line) on the vertical axis and displays the independent variable on the horizontal axis.  A linear regression model becomes appropriate for a dataset when the points are randomly dispersed around the horizontal axis near 0; otherwise, a nonlinear model becomes more appropriate.

Answer:

x = 2 yd

Step-by-step explanation:

Angles of 45 degreees = two congruent legs

for the Pythagorean theorem

2x^2 = 8

x^2 = 4

x = 2

9514 1404 393

Answer:

  152.3°

Step-by-step explanation:

The arcsine function only gives angles in quadrants I and IV. Since this is a quadrant II angle, its value will be ...

  θ = 180° -arcsin(0.4649) = 180° -27.7°

  θ = 152.3°

Answer:

6'2

Step-by-step explanation:

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: I think the answer is A

Step-by-step explanation:

Answer:

6/25

Step-by-step explanation:

P(have belief) =76/100 = 19/25

P (does not have belief) = 1-19/25 = 6/25

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]        

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]