pls can solve this question . thanks

[tex]4x + 18 = 7x - 21 \\ 18 + 21 = 7x - 4x \\ 39 = 3x \\ x = \frac{39}{3} = 13 [/tex]

The matrix [tex]\vec C[/tex] resulting from the multiplication between matrices [tex]\vec B[/tex] and [tex]\vec A[/tex] is equal to the matrix [tex]\left[\begin{array}{ccc}37&55&- 20\\23&11&- 20\\25&28&-9\end{array}\right][/tex]. (Correct choice: C)

In this question we must apply the operation of multiplication of two matrices, whose definition is shown below:

[tex]\vec B\, \cdot \, \vec A[/tex], where [tex]\vec B \in \mathbb {R}_{n \times p}[/tex] and [tex]\vec A \in \mathbb{R}_{p \times n}[/tex]     (1)

Where each element of the resulting matrix is equal to the following expression:

[tex]c_{(i, j)} = b_{(i, p)} \,\bullet\, a_{(p, j)}[/tex]     (2)

Where the operator "[tex]\bullet[/tex]" represents a dot product.

Now we proceed to calculate each element of the matrix:

[tex]c_{(1, 1)} = \left[\begin{array}{ccc}5&7&3\end{array}\right] \,\bullet\,\left[\begin{array}{c}1\\5\\- 1\end{array}\right][/tex]

5 · 1 + 7 · 5 + 3 · (- 1)

37

[tex]c_{(1, 2)} = \left[\begin{array}{ccc}5&7&3\end{array}\right] \,\bullet\,\left[\begin{array}{c}7\\2\\2\end{array}\right][/tex]

5 · 7 + 7 · 2 + 3 · 2

55

[tex]c_{(1, 3)} = \left[\begin{array}{ccc}5&7&3\end{array}\right] \,\bullet\,\left[\begin{array}{c}- 3\\1\\- 4\end{array}\right][/tex]

5 · (- 3) + 7 · 1 + 3 · (- 4)

- 20

[tex]c_{(2, 1)} = \left[\begin{array}{ccc}- 3&7&9\end{array}\right] \,\bullet\,\left[\begin{array}{c}1\\5\\- 1\end{array}\right][/tex]

(- 3) · 1 + 7 · 5 + 9 · (- 1)

23

[tex]c_{(2, 2)} = \left[\begin{array}{ccc}- 3&7&9\end{array}\right] \,\bullet\,\left[\begin{array}{c}7\\2\\2\end{array}\right][/tex]

(- 3) · 7 + 7 · 2 + 9 · 2

11

[tex]c_{(2, 3)} = \left[\begin{array}{ccc}- 3&7&9\end{array}\right] \,\bullet\,\left[\begin{array}{c}- 3\\1\\- 4\end{array}\right][/tex]

(- 3) · (- 3) + 7 · 1 + 9 · (- 4)

- 20

[tex]c_{(3, 1)} = \left[\begin{array}{ccc}2&5&2\end{array}\right] \,\bullet\,\left[\begin{array}{c}1\\5\\- 1\end{array}\right][/tex]

2 · 1 + 5 · 5 + 2 · (- 1)

25

[tex]c_{(3, 2)} = \left[\begin{array}{ccc}2&5&2\end{array}\right] \,\bullet\,\left[\begin{array}{c}7\\2\\2\end{array}\right][/tex]

2 · 7 + 5 · 2 + 2 · 2

28

[tex]c_{(3, 3)} = \left[\begin{array}{ccc}2&5&2\end{array}\right] \,\bullet\,\left[\begin{array}{c}- 3\\1\\- 4\end{array}\right][/tex]

2 · (- 3) + 5 · 1 + 2 · (- 4)

- 9

The matrix [tex]\vec C[/tex] resulting from the multiplication between matrices [tex]\vec B[/tex] and [tex]\vec A[/tex] is equal to the matrix [tex]\left[\begin{array}{ccc}37&55&- 20\\23&11&- 20\\25&28&-9\end{array}\right][/tex]. (Correct choice: C)

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Answer:

road

Step-by-step explanation:

Answer:

235÷18=

how many classes of 18 students for each teacher

Step-by-step explanation:

Answer:switch mikes Luke’s and Alex in between each other and let winter keep her own

Step-by-step explanation:

Given :

[tex] \: [/tex]

[tex] \: [/tex]

[tex] \gray{ \frak{The \: given \: two \: \: points \: are(x_{1 },y_{1})=(−6 , -10)and(x_{ 2 },y_{2} )=( 2,5 )\:}}[/tex]

[tex] \: [/tex]

Let's solve by using midpoint formula :

[tex] \: [/tex]

[tex] \bf \boxed{\color{red}\frak{Midpoint \: \: Formula : {( \: x ,\:y \: ) = }( \frak{ \frac{x_{1 } + y_{1}}{2} , \frac{x_{2 } + y_{2}}{2} )}}}[/tex]

[tex] \: [/tex]

[tex] \large \gray{ \frak{( \: x ,\:y \: ) = }( \frak{ \frac{ -6 + 2}{2} , \frac{ - 10 + 5}{2} )}}[/tex]

[tex] \: [/tex]

[tex] \: \large \gray{ \frak{( \: x ,\:y \: ) = }( \frak{ \frac{ -4}{2} , \frac{ - 5}{2} )}}[/tex]

[tex] \: [/tex]

[tex]\large \gray{ \frak{( \: x ,\:y \: ) = }( \frak{ \cancel\frac{ -4}{2} , \cancel\frac{ - 5}{2} )}}[/tex]

[tex] \: \: [/tex]

[tex] \underline{ \boxed{ \large \red{ \frak{option \: d }( \frak{ - 2, - 2.5)}}}}✓[/tex]

Hope Helps! :)

Answer: False

Step-by-step explanation:

The only number that makes this impossible to solve is x = 0

Answer:

no

Step-by-step explanation:

because no

Answer:7 people

Step-by-step explanation:

Answer:

7

Step-by-step explanation:

It is an example of Type I error.

About Type I error

The rejection of a null hypothesis (H0) when it is actually true is a Type I error. It is represented by the following condition,

Failure to reject a null hypothesis when it is true that it is false which is a Type II error.

How is this an example of Type I error?

The given case's hypothesis is described as follows:

H₀: The product has no effect on the plant's height.

Hₐ: The substance causes the plant to enlarge.

The sample claims that the product causes the plant to grow taller, however in reality, it has no effect on the plant's height.

Therefore, the sample recommends rejecting the null hypothesis even though it is true. Consequently, a result, this error is a Type I error.

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Answer:

I have attached a really rough graph!!

Step-by-step explanation:

The point that does not fit in the proportional relationship is (6,2) when we plot the other three points and join the points they will give us a staright line but the point (6,2) isnt lying on that staright line.. Therefore, it does not fit in the proportional relationship!

Answer:

(6,2)

Step-by-step explanation:

Answer: -2 and 4

Step-by-step explanation:

Let [tex]t=x/2[/tex].

[tex]f(t)=0 \longrightarrow t=-1, 2\\\\\therefore x=-2, 4[/tex]

Using the hypergeometric distribution, the probabilities are given as follows:

a) 0.2 = 20%.

b) 0.5539 = 55.39%.

The formula is:

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

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

Researching this problem on the internet, it is found that the council has 15 members, of which 7 are Democrats and 8 are Republicans.

Item a:

The parameters are:

N = 15, n = 2, k = 7.

The probability is P(X = 2), hence:

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

[tex]P(X = 2) = h(2,15,2,7) = \frac{C_{7,2}C_{8,0}}{C_{15,2}} = 0.2[/tex]

Item b:

At most one Democrat, with n = 3, hence:

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

In which:

[tex]P(X = 0) = h(0,15,3,7) = \frac{C_{3,0}C_{8,3}}{C_{15,3}} = 0.1231[/tex]

[tex]P(X = 1) = h(1,15,3,7) = \frac{C_{3,1}C_{8,2}}{C_{15,3}} = 0.4308[/tex]

Then:

[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1231 + 0.4308 = 0.5539[/tex]

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Answer: Look at step by step

Step-by-step explanation:

(a) There are 119 stocks that increased in price out of 246 stocks so the probability of selecting a stock that is increased in price is 119/246 = 48.37%

(b) There are 112 stocks where it pays dividends so the probability so the probability of selecting a stock that pays dividends is 112/246 = 45.53%

(c) There are 49 stocks which satisfy the question so the probability is 49/246 = 19.92%

(e) There are 34 stocks which increase in price and pay dividends out of the 119 stocks that increase in price so the probability is 34/119 = 28.57%

(f) Using the same logic as (e), the probability is 85/134 = 63.43%

(g) From the requirements, we see that 49 stocks don't satisfy the requirements, so the probability is (246-49)/246 = 80.08%

The 1st table represents a linear as well as the quadratic relationship, the 2nd table represents neither and the 3rd table also represents neither relationship.

Hence, the 1st table represents a linear as well as a quadratic relationship, the 2nd table represents neither and the 3rd table also represents neither relationship.

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The current stock price is $54.15.

Given that  RLX Company just paid a dividend of $2.85 per share on its stock and dividends are expected to grow at a constant rate of 4.5 percent per year, indefinitely.

Stock valuation alludes to the valuation of the natural worth per portion of an organization's stock which can help the organization in deciding its reasonable worth in case of twisting up or consolidation. One of the models used to esteem the stock cost is the profit development model. It expects the cost of the stock to be the current worth representing things to come profits of the stock accepted to develop at a consistent rate.

Dividend (D0) = $2.85

Growth rate (g) = 4.5%

Required return (r) = 10%

Firstly, we have to calculate the value of the current stock price.

The current stock price can be calculated by using the dividend growing model.

[tex]\begin{aligned}\text{Current stock price}&=\frac{D_{0}\times(1+g)}{r-g}\\ &=\frac{2.85\times (1+0.045)}{0.10-0.045}\\ &=\frac{2.85\times 1.045}{0.055}\\ &=\frac{2.97825}{0.055}\\ &=54.15\end[/tex]

Hence, the current stock price when dividend of $2.85 are expected to grow at a constant rate of 4.5 percent per year, indefinitely is $54.15.

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Answer:

the rectangles area is 24 meters

Step-by-step explanation:

area = length x width

length is 6 using the words long and width is 4

Answer:

area = 24 cm²

Step-by-step explanation:

The area of a rectangle can be found using the formula:

[tex]\boxed {area = length \times width}[/tex].

Substituting the values into the equation:

[tex]area = 6 \space\ cm\times 4 \space\ cm[/tex]

       ⇒ [tex]\bf 24 \space\ cm^2[/tex]

Answer:

1.0 x 10^0

Step-by-step explanation:

The above works because 10^0 = 1, and 1 x 1 = 1.

Brainliest, please :)

The solutions to the system of equations are (4, 15) and (-1,-1)

The system of equations is given as:

y=3x+2

y=x^2-4+2

Substitute y=3x+2 in y=x^2-4+2

3x + 2=x^2-4+2

Simplify

x^2 - 3x - 4 = 0

Expand

x^2 + x - 4x - 4x = 0

Factorize

x(x + 1) - 4(x + 1) = 0

This gives

(x - 4)(x + 1) = 0

Solve for x

x = 4 and x = -1

Substitute these values in y=3x+2

y = 3(4)+2 = 15

y = 3(-1)+2 = -1

Hence, the solutions to the system of equations are (4, 15) and (-1,-1)

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The largest size of a team according to the task content is; 20 kids.

It follows from the task content that the population is; 20 students in Mrs. Tanya's class and 24 kids in Mrs. Olga's class.

However, since, the team forming condition states that students be placed on teams of equal size with students from only their own class.

It therefore follows that the largest size of a team is 20 as only 20 students are in Mrs. Tanya's class hence leaving 4 kids in Mrs. Olga's class without a team.

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Answer:

A

Step-by-step explanation:

the sum to n terms of a geometric series is

[tex]S_{n}[/tex] = [tex]\frac{a_{1}(r^{n}-1) }{r-1}[/tex] , then for n = 7

S₇ = [tex]\frac{-1(2^{7}-1) }{2-1}[/tex] = [tex]\frac{-1(128-1)}{1}[/tex] = - 1 (127) = - 127

Using a system of equations, it is found that there were 100 students at the bus stop.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

Considering the first paragraph, the equation is:

[tex]\frac{x}{y - 25} = \frac{3}{2}[/tex]

Considering the second paragraph, the equation is:

y - 25 = x - 15.

Hence:

y = x + 10.

Replacing on the first equation:

[tex]\frac{x}{x - 15} = \frac{3}{2}[/tex]

3x - 45 = 2x

x = 45.

y = x + 10 = 55.

The total number was:

45 + 55 = 100.

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Using the monthly payment formula, he needs to deposit A = $1,342.12 a month to reach his goal.

It is given by:

[tex]A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}[/tex]

In which:

For this problem, the parameters are:

P = 250000, r = 0.05, n = 12 x 30 = 360.

Then:

r/12 = 0.05/12 = 0.004167.

Then:

[tex]A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}[/tex]

[tex]A = 250000\frac{0.004167(1+0.004167)^{360}}{(1+0.004167)^{360} - 1}[/tex]

A = $1,342.12

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The doubling time after 4 and 7 years is 400 months and 700 months respectively

The formula for finding doubling time is given as;

Doubling time = [tex]t \frac{In 2}{In 1 + \frac{r}{100} }[/tex]

Where t = time = 4 and 7 years

r = rate

For 4 years, we have

Doubling time = [tex]4 * \frac{In 2}{In ( 1 + \frac{0.7}{100}) }[/tex]

⇒ [tex]4 * \frac{In 2}{In 1. 0074}[/tex]

⇒ [tex]4 * \frac{0. 6931}{0. 0074}[/tex]

⇒ [tex]4[/tex] × [tex]93. 66[/tex]

⇒ 374. 65

⇒ 400 months in the nearest hundredth

For 7 years

⇒ [tex]7[/tex] × [tex]93. 66[/tex]

⇒ [tex]655. 62[/tex]

⇒ 700 months in the nearest hundredth

Thus, the doubling time after 4 and 7 years is 400 months and 700 months respectively

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Answer:

the real answer is 2 I hope that

The number of board game sorted is 14

The games have been sorted into video games and board games.

The number of video games is 28 less than 6 times the number of board games.

Therefore,

the number of board games = x

number of video game = 6x - 28

The games are 80% video games. Hence,

80 / 100 × x + 6x - 28 = 6x - 28

4 / 5 × 7x - 28 = 6x - 28

4 / 5 (7x - 28) = 6x -28

28 / 5 x - 112 / 5  = 6x - 28

28 / 5 x  - 6x = -28 + 112 / 5

28x - 30x / 5 =  - 28 + 112 / 5

-2x / 5  = -28 / 5

-10x = -140

x = 14

Therefore, the number of board game sorted is 14.

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Answer:

14

Step-by-step explanation:

80 / 100 × x + 6x - 28 = 6x - 28

4 / 5 × 7x - 28 = 6x - 28

4 / 5 (7x - 28) = 6x -28

28 / 5 x - 112 / 5  = 6x - 28

28 / 5 x  - 6x = -28 + 112 / 5

28x - 30x / 5 =  - 28 + 112 / 5

-2x / 5  = -28 / 5

-10x = -140

x = 14

Answer:

log(a^x^2-27(radicalx))

Step-by-step explanation:

Assuming you have written your question correctly, this is it.

x=regular selling price

X×80/100=380

= X×8/10=380

= 8X=380×10=3800

X=3800/8

X=475

the regular selling price is 475

Before the 20% discount, the TV's regular selling price was $475 as per the concept of percentage.

Let's denote the regular selling price of the plasma TV as "x". When the TV was marked down by 20%, the sale price became $380.

To find the regular selling price, we can set up an equation using the information given:

Sale Price = Regular Selling Price - (Discount Percentage × Regular Selling Price)

$380 = x - (0.20x)

Now, let's solve for "x" (the regular selling price):

$380 = x - 0.20x

$380 = 0.80x

To isolate "x," we divide both sides of the equation by 0.80:

x = $380 / 0.80

x = $475

The regular selling price of the plasma TV is $475.

Therefore, before the 20% discount, the TV's regular selling price was $475. After the discount, the sale price was reduced to $380, which is 20% less than the regular price.

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