What is the solution to this system of equations?
X+ 2y = 4
2x-2y = 5
0 (3.-52
0 (3.3
O no solution
infinitely many solutions

By using the multiple regression equation, the predicted y-values for the values of the given independent variables include the following:

(a) ý = 207.02.

(b) ý = 194.18.

(c) ý = 270.02.

(d) ý = 305.42.

In Statistics and Mathematics, a regression line simply refers to a statistical line that best describes the behavior of a data set. This ultimately implies that, a regression line refers to a line which best fits a set of data.

Next, we would determine the predicted y-values for the values of the given independent variables as follows;

ý = -51.7+03x₁ + 4.8x₂

(a) x₁ = 72 and x₂ = 8.9

ý = -51.7+03(72) + 4.8(8.9)

ý = 207.02.

(b) x₁ = 65 and x₂ = 10.6

ý = -51.7+03(65) + 4.8(8.9)

ý = 194.18.

(c) x₁ = 81 and x₂ = 16.4

ý = -51.7+03(81) + 4.8(16.4)

ý = 270.02.

(d) x₁ = 88 and x₂ = 19.4

ý = -51.7+03(88) + 4.8(19.4)

ý = 305.42.

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Answer:

60π cm2

Step-by-step explanation:

Interval of increase: (-∞, -1) ∪ (3, +∞)

Interval of decrease: (-1, 3)

Given function is f(x) = x³ - 3x² - 9x + 4

Differentiating with respect to x,

f'(x) = 3x² - 6x - 9

f'(x) = 3 (x² - 2x - 3)

f'(x) = 3 (x² - 3x + x - 3)

f'(x) = 3 (x(x - 3) + (x - 3))

f'(x) = 3 (x - 3) (x + 1)

For increasing or decreasing,

f'(x) = 0

3 (x - 3) (x + 1) = 0

x = - 1, 3

In interval (-∞, -1 ), f'(x) = positive,

⇒ f(x) is increasing on (-∞, -1 ),

In interval (-1, 3), f'(x) = negative,

⇒ f(x) is decreasing on (-1, 3),

In interval (3, ∞), f'(x) = positive,

⇒ f(x) is increasing on (3, ∞),

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Given question is incomplete, the complete question is below

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x³ - 3x² - 9x + 4 (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.)

Answer:

[tex]the \: angle \: is \: 0.25\pi[/tex]

Step-by-step explanation:

the arc length is equal to the radius x angle (in radians)

[tex]s=ra[/tex]

s= arc length, r= radius and a= the angle

solve for a ; a= s/r; a=

[tex]1.25\pi/5[/tex]

The measure of the arc is 0.25π.

Important information:

The formula for arc length is:

[tex]s=r\theta[/tex]

where s is the arc length, r is the radius and [tex]\theta [/tex] is the measure of arc in radian.

The diameter of a circle is 10 inches. So, the radius is 5 inches.

[tex]1.25\pi=5\theta[/tex]

[tex]\dfrac{1.25\pi}{5}=\theta[/tex]

[tex]0.25\pi=\theta[/tex]

Therefore, the measure of the arc is 0.25π.

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Answer:

42 square units

Step-by-step explanation:

Each square on the grid is 2 units by 2 units.

Bases: OC = 10; SK = 4

Height: OS = 6

area = (b1 + b2)h/2

area = (10 + 4)(6)/2

area = 42

Answer: 42 square units

Answer:

The area of trapezoid is 42 square unit.

Step-by-step explanation:

According to the question

A trapezoid which is consist of many square shapes and each grid square equals 2 units . We need to find the area of trapezoid.

To calculate the area of trapezoid we have given :-

Finding the area of trapezoid :-

Area of trapezoid = ½ × ( 4 + 10 ) × 6

Area of trapezoid = ½ × 14 × 6

Area of trapezoid = ½ × 84

Area of trapezoid = 42

Hence,

The area of trapezoid is 42 square unit.

The apparent capacity of a glass vessel refers to its perceived size or volume based on external dimensions, while the actual capacity represents the true volume of the vessel, taking into account both interior and exterior dimensions.

[tex]\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}[/tex]

♥️ [tex]\large{\textcolor{red}{\underline{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]

Answer:

72[tex]\sqrt{3}[/tex]

Step-by-step explanation:

The area (A) of a rhombus is calculated as

A = [tex]\frac{1}{2}[/tex] × d₁ × d₂ (d₁ and d₂ are the diagonals )

The diagonals bisect each other at right angles

d₁ = 2 × 6 = 12

Use the tangent ration in the upper left right triangle and the exact value

tan60° = [tex]\sqrt{3}[/tex]

tan60° = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{opp}{6}[/tex] = [tex]\sqrt{3}[/tex] ( multiply both sides by 6 )

opp = 6[tex]\sqrt{3}[/tex] , then

d₂ = 2 × 6[tex]\sqrt{3}[/tex] = 12[tex]\sqrt{3}[/tex]

Thus

A = [tex]\frac{1}{2}[/tex] × 12 × 12[tex]\sqrt{3}[/tex] = 6 × 12[tex]\sqrt{3}[/tex] = 72[tex]\sqrt{3}[/tex]

Answer:

52 years

Step-by-step explanation:

Let

Your age = x

3 times your age = 3 * x

= 3x

The equation can be written as:

400 - 3x = 244

400 - 244 = 3x

156 = 3x

Divide both sides by 3

x = 156/3

x = 52

Therefore,

Your age is 52 years

Answer:

The rate of change or slope is 2

Step-by-step explanation:

Rise over run 2/1

Answer:

1.

Mean:22

Median: 19.5

Mode:12

Range:26

2.

Mean: 21.143

Median: 16

Mode: 5

Range: 43

3.

Mean: 38.5

Median: 37.5

Mode: 13 , 34, 41

Range: 56

4.

Mean: 73.167

Median: 78.5

Mode: 85

Range : 47

5.

Mean: 58.75

Median: 63

Mode: 63

Range : 70

6.

Mean: 68.714

Median: 74

Mode: 74

Range: 57

7.

Mean: 48

47.5

72

66

8.

59.625

61

90

79

9.

31.571

32

46

34

10.

42.111

40

51

51

Answer:12 : 6 22 : 11 32 : 16

Step-by-step explanation:

Answer:

4:2

Step-by-step explanation:

Answer:

lim [(x(2) + 3) × 1/ x(4) ] = 3 × ( + ○○) = + ○○

Answer:

The age of paintings is - 26,601 years .

Step-by-step explanation:

Lets calculate-

A= [tex]0.25 A[/tex]o

Therefore ,

[tex]A= A[/tex]o [tex]e^-^0^.^0^0^0^1^2^1^t[/tex]

[tex]ln0.25=-0.000121t[/tex]

[tex]-3.2188=-0.000121t[/tex]

[tex]t=\frac{-3.2188}{-0.000121}[/tex]

  t= 26,601.65

Therefore , the age of paintings is 26,601 years old.

The correct hypotheses to test is (f) H₀ : μ₁ = μ₂; H₁ : μ₁ < μ₂

The test statistic is undefined

The critical value at α = 0.01 is 2.58

From the question, we have the following parameters that can be used in our computation:

Mean time served for fraud is less than that for firearms offenses

This means that the hypotheses to test are

H₀ : μ₁ = μ₂

H₁ : μ₁ < μ₂

This is represented with option (f)

The test statistic can be calculated using

z = (x₁/n₁ - x₂/n₂)/√[p(1 - p)/n₁ + p(1 - p)/n₂)

Where p = pooled sample proportion

and p = (x₁ + x₂)/(n₁ + n₂)

The required values are not given

So, the test statistic is undefined

Given that

α = 1%

This means that

α = 0.01

The critical value at α = 0.01 is 2.58

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Answer:

the answer is 20 because you round of 21

A random variable is normally distributed with a mean of 40 and a standard deviation of 2.5. P(XS 43.75) is  0.9332

Given: A random variable is normally distributed with a mean of 40 and a standard deviation of 2.5.

We have to find P(X<43.75).

We have to find the probability of getting a random variable below 43.75 using the normal distribution formula.

Z = (X - μ) / σ

Z = (43.75 - 40) / 2.5

Z = 1.5

The probability of a random variable being less than 43.75 is the same as the probability of a standard normal random variable being less than 1.5.

P(Z < 1.5) = 0.9332 (using the standard normal distribution table)

Thus, the P(X < 43.75) is approximately 0.9332 (to four decimal places).

Hence, the required probability is 0.9332.

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1) the null and alterative hypotheses are  - H_o: p=0.20 H_1:p<0.26 (option C)

2)

1) The null hypothesis is that the population proportion of yellow peas is equal to 20%. The alternative hypothesis is that the population proportion of yellow peas is less than 20%.

hence,  H_o: p=0.20 H_1:p<0.26

2) The test statistic is calculated as follows  -

z = (pa   - p₀) / √(p₀(1-p₀)/ n)

Where  -

pa is the sample proportion of yellow peas (0.25)

p₀ is the hypothesized proportion of yellow peas (0.20)

n is the sample size (537)

The test statistic is -2.17

p- value =P(Z < -2.17)

The p -value is   0.0306.

Since the p-value is less than the significance level of 0.05,we reject the null hypothesis and    conclude that there is sufficient evidence to suggest that the population proportion of yellow peas is less than 20%.

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Answer:

E. 16

Step-by-step explanation:

Easy:

Formula: P=2(l+w)

Plug in and solve:

P=2(3+5)

P = 16 (Option E is correct)

Answer:

well to tell you its easy 4 me

Given:

Leading coefficient of a trinomial = 3

Constant term = [tex]-5[/tex]

To find:

The equation of trinomial.

Solution:

If an expression contains three terms, then it is known as trinomial.

The general form of a trinomial which contains leading term and a constant is:

[tex]ax^2+bx+c[/tex]

Where, a,b,c are real non-zero numbers. Here, a is the leading coefficient and c is the constant.

Putting a=3 and c=-5, we get

[tex]3x^2+bx+(-5)[/tex]

[tex]3x^2+bx-5[/tex]

Therefore, the required trinomial is  [tex]3x^2+bx-5[/tex], where b is a non zero real number.

Answer:

[tex] \sqrt{12a \: + \: 12b \: - \: 2ab \: + \: 2b {}^{2} } [/tex]

Step-by-step explanation:

(a²-b²) / (a+b) = 12

a² - b² = 12(a+b)

(a-b)² + 2ab - 2b² = 12a + 12b

(a-b)² = 12a + 12b - 2ab + 2b²

a - b =

[tex] \sqrt{12a \: + 12b \: - 2ab \: + \: 2b {}^{2} } [/tex]

Answer:

y=5x-13

Step-by-step explanation:

The graph of the functions fo, f1, and f2 can be sketched as follows: fo(x) is a constant function equal to 1 on the interval [0, 2]. f1(x) is a piecewise function that equals x³ + 1 on [0, 1] and 1 on (1, 2]. f2(x) is a piecewise function that equals 1 on [0, 1] and 2 on (1, 2].

To sketch the graph of the functions fo, f1, and f2, we consider their definitions and the given intervals.

The function fo(x) is a constant function that equals 1 on the entire interval [0, 2]. Therefore, its graph will be a horizontal line at y = 1 throughout the interval.

The function f1(x) is a piecewise function. On the interval [0, 1], it equals x³ + 1, which means it starts at y = 1 when x = 0 and increases as x moves towards 1. At x = 1, there is a jump discontinuity, and the function becomes a constant 1 for x in (1, 2]. Therefore, the graph of f1 will be a curve increasing from y = 1 to some maximum point between x = 0 and x = 1, and then it remains constant at y = 1 for x in (1, 2].

The function f2(x) is also a piecewise function. On the interval [0, 1], it equals 1, so its graph will be a horizontal line at y = 1. At x = 1, there is another jump discontinuity, and the function becomes a constant 2 for x in (1, 2]. Therefore, the graph of f2 will have a horizontal line at y = 1 on [0, 1] and a horizontal line at y = 2 on (1, 2].

By considering the definitions and intervals, we can sketch the graphs of fo, f1, and f2 as described above.

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