MN=3X+2 AND NP=X+4 WHAT IS THE VALUE OF X

Answer:

x = 5

Step-by-step explanation:

To solve the equation 4x-8 + 4 = 16, we first combine the constants on the left-hand side:

4x-8+4 = 16

4x-4 = 16

Next, we add 4 to both sides to isolate the variable on one side:

4x-4+4 = 16+4

4x = 20

Finally, we divide both sides by 4 to solve for x:

4x/4 = 20/4

x = 5

Therefore, the solution to the equation 4x-8 + 4 = 16 is x = 5.

Answer:

1. # of people who predicted they would pass = 30

2. # of people who predicted they would fail = 20

3. # of people who predicted they would pass and actually passed = 27

4. # of people who predicted they would pass and actually failed = 3

5. # of people who predicted they would fail and actually passed = 11

6. # of people who predicted they would fail and actually failed = 9

Step-by-step explanation:

1. # of people who predicted they would

I also filled in the frequency table by extracting it from Brainly and drawing on it to show how the math works and fits in the table.

Using the equations given, only equation 2 has a proportional relationship.

Proportional relationships are relationships between two variables where their ratios are equivalent. Another way to think about them is that, in a proportional relationship, one variable is always a constant value times the other.

The difference between proportional and non-proportional linear relationships is that, although both have constant rates of change the proportional relationship has constant output to input ratios, and the non-proportional relationship does not.

In this problem, to determine the proportional relationship, we can also check for both values of x and y.

1. y = -0.75x + 5

When x = 1

y = -0.75(1) + 5

y = 4.25

when x = 2

y = -0.75(2) + 5

y = 3.5

The relationship is not proportional since as one variable increase, the other decreases.

2. y = 4x - 1

When x = 1

y = 4(1) - 1

y = 4 - 1

y = 3

when x = 2

y = 4(2) - 1

y = 8 - 1

y = 7

This relationship is proportional since the variables increases or decreases with one another.

3. y = (3/4)x

When x = 1

y = (3/4)(1)

y = 3/4

when x = 2

y = (3/4)(2)

y = 2/3

This relationship is not proportional

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Answer:

Words of the Devil, by Parau na te Varua ino, is a painting that was created in 1892 in French Polynesia. The painting depicts a scene from Tahitian mythology, in which the goddess Hina is visited by the devil, who is represented as a horned figure with a large, forked tongue. The painting is rich in detail and vivid color, with the figures depicted in bright, bold colors that stand out against the darker background.

The style of the painting is characteristic of the Tahitian art of the time, which was heavily influenced by the traditional Polynesian art forms of carving, tattooing, and tapa cloth decoration. The figures in the painting are depicted in a stylized, almost abstract way, with exaggerated features and bold, sweeping lines. The use of bright colors and bold brushstrokes gives the painting a dynamic, almost chaotic energy that reflects the intense emotions of the scene.

Overall, Words of the Devil is a powerful and striking work of art that captures the mythology and cultural traditions of French Polynesia in the late 19th century.

13.) The simplified polynomials that can represent the area and perimeter of the large rectangle =40x + 5x² and

12X + 16 respectively.

14.)The simplified polynomials that can represent the area and perimeter of the large rectangle =21+10x + x² and 20+4x respectively.

To determine the simplified polynomials, the formula for the area and perimeter of the rectangule should be used. That is:

Area of rectangle = length×width

where;

length = 5x

width = 8+X

Area = 5x * 8+X

= 40x + 5x²

Perimeter of rectangle = 2(length+width)

= 2(5x + 8+X)

= 10x + 16 + 2x

= 12X + 16

For question 14.)

Area of rectangle = length× width

where;

length = 7+X

width = 3+X

Area = 7+X × 3+X

= 21+7x+3x +x²

= 21+10x + x²

Perimeter = 2(length+width)

= 2(7+X+3 + X)

= 2(10+2x)

= 20+4x

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The amount you will have in the account in 30 years is $22836.77

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

Using the above as a guide, we have the following:

Amount = P * (1 + r)^t

Where

P = Principal = 3000

r = Rate = 7%

t = time = 30

Substitute the known values in the above equation, so, we have the following representation

Amount = 3000 * (1 + 7%)^30

Evaluate

Amount = 22836.77

Hence, the amount is 22836.77

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The reading that separates the rejected thermometers from the others is given as follows:

1.175 ºC.

The z-score of a measure X of a normally distributed variable that has mean represented by [tex]\mu[/tex] and standard deviation represented by [tex]\sigma[/tex] is obtained by the equation presented as follows:

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

The mean and the standard deviation for this problem are given as follows:

[tex]\mu = 0, \sigma = 1[/tex]

The 12% higher of temperatures are rejected, hence the 88th percentile is the value of interest, which is X when Z = 1.175.

Hence:

1.175 = X/1

X = 1.175 ºC.

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The inequality in standard form that will represent the maximum number of cans of apple juice and cans of pineapple juice Stewart can use would be = 240ounce = 10a + 18p.

The total quantity of fruit punch that would be used for the birthday = 240 ounce.

The quantity of juice each can holds for apple juice,a, = 10 ounces.

The quantity of juice each can hold for pineapple juice,p,= 18 ounces.

Therefore the best expression in standard form would be = 240ounce = 10a + 18p.

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Answer:

  54

Step-by-step explanation:

You want the approximate test grade of a student with a homework grade of 61, using a linear regression equation for the given homework and test scores.

The linear regression equation can be found using any of a number of calculators, apps, or spreadsheets. The attachment shows the equation is approximately ...

  y ≈ 1.22x -20.01

For a homework score of 61, the expected test grade is ...

  y = 1.22(61) -20.01

  y ≈ 54

The student's projected test grade is 54.

__

Additional comment

If the regression coefficients are used to their full precision, the projected test score is about 54.6, which rounds to 55.

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When we evaluate the given expression, A/5 + √(B - C), the result obtained is 9 (option B)

First, we shall determine the value of A. Details below:

Quadratic equation is expressed as:

x² - (sum of root)x + product of root

Comparing the above with x² - 11x + 30, we have

x² - 11x + 30 = x² - (sumof root)x + product of root

Product of roots = 30

Thus,

A = 30

Next, we shall determine the value of B. details below:

f(x) = x² + 5

f(2) = 2² + 5

f(2) = 9

Thus,

B = 9

Next, we shall determine the value of C. Details below:

Let

x + 4 = 0

Thus,

x = -4

Substitute the value of x into x² - 2x - 24 to obtain the remainder as shown below:

Remainder = x² - 2x - 24

Remainder = (-4)² - 2(-4) - 24

Remainder = 0

Thus,

C = 0

Finally, we shall determine value of A/5 + √(B - C). Details below:

A/5 + √(B - C) = 30/5 + √(9 - 0)

A/5 + √(B - C) = 6 + √(9

A/5 + √(B - C) = 6 + 3

A/5 + √(B - C) = 9

Thus, the value of A/5 + √(B - C) is 9 (option B)

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Answer:

A = 40 cm^2

Step-by-step explanation:

The area of a triangle is given by

A = 1/2 bh where b is the base and h is the height

A = 1/2 ( 20) (4)

A = 40 cm^2

Answer:

40in²

Step-by-step explanation:

The formula to find the area of a triangle is:

[tex]\sf Area = \frac{1}{2}*Base*Height[/tex]

Accordingly, let us solve it now.

[tex]\sf \frac{1}{2}*Base*Height\\\\\sf \frac{1}{2}*20*4\\\\40\:in^2[/tex]

after 1000 years, there will be approximately 218.7 mg of Radium-226 remaining in the sample. use formula for radioactive decay to solve

what is radioactive decay ?

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation, such as alpha particles, beta particles, and gamma rays. The decay occurs in order to achieve a more stable configuration of the nucleus.

In the given question,

The formula for radioactive decay is:

N(t) = N0 * (1/2)ᵃ⁾ᵇ

where:

N(t) is the amount of radioactive material at time t

N0 is the initial amount of radioactive material

t is the time elapsed since the initial measurement

T is the half-life of the radioactive material

We can use this formula to solve the problem as follows:

N0 = 500 mg (the initial amount)

T = 1590 years (the half-life)

t = 1000 years (the time elapsed)

N(t) = 500 * (1/2)¹⁰⁰⁰⁾ ¹⁵⁹⁰

N(t) = 500 * 0.4374

N(t) = 218.7 mg

Therefore, after 1000 years, there will be approximately 218.7 mg of Radium-226 remaining in the sample.

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Answer:

So when the two vehicles meet, the van has travelled 87.5 km and the car has travelled 150 km.

Step-by-step explanation:

(a) Let's call the time it takes for the two vehicles to meet "t". We know that the distance between the two towns is 237.5 km, and the combined speed of the two vehicles is 35 km/h + 60 km/h = 95 km/h. Using the formula distance = speed × time:

237.5 = 95t

Solving for t:

t = 237.5/95

t ≈ 2.5 hours

So the two vehicles will meet on the way 2.5 hours after 11.30 a.m., which is at 2.00 p.m.

(b) To find how far each vehicle has traveled when they meet, we can use the formula distance = speed × time again. The van travels at 35 km/h for 2.5 hours, so it travels:

distance = speed × time = 35 km/h × 2.5 hours = 87.5 km

The car travels at 60 km/h for 2.5 hours, so it travels:

distance = speed × time = 60 km/h × 2.5 hours = 150 km

So when the two vehicles meet, the van has traveled 87.5 km and the car has traveled 150 km.

The resulting matrix formed by performing R2 -> 4R1 + R2 on M is given as follows:

[tex]M = \left[\begin{array}{ccc}-4&3&1\\-18&9&8\end{array}\right][/tex]

The matrix in the context of this problem is defined as follows:

[tex]M = \left[\begin{array}{ccc}-4&3&1\\-2&-3&r\end{array}\right][/tex]

The rows of the matrix are given as follows:

Hence the row 2 of the resulting matrix has the elements given as follows:

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Answer:

The answer is A

Step-by-step explanation:

Trust me

A: The data provides convincing evidence, at α=0.02 level, that the glucose drug is effective in reducing mean glucose level.

B: The 98% confidence interval for the difference in mean reduction of glucose level between placebo and drug groups is 3.5 to 13.7 mg/dL, supporting the effectiveness of the glucose drug

A: To test whether the glucose drug is effective in producing a reduction in mean glucose level, we will use a two-sample t-test with equal variances assuming normality of the differences.

Let μA be the true mean reduction in glucose level for the placebo group and μB be the true mean reduction in glucose level for the glucose drug group. The null hypothesis is H0: μA - μB = 0, and the alternative hypothesis is Ha: μA - μB < 0

Using the given data, the sample mean reduction for the placebo group is 9.7 mg/dL and for the glucose drug group is 18.3 mg/dL. The pooled sample standard deviation is 8.064 mg/dL, and the t-statistic is calculated to be:

t = (xB - xA) / (sP × √(1/nA + 1/nB))

= (18.3 - 9.7) / (8.064 × √(1/10 + 1/10))

= 2.551

where xA and xB are the sample means, sP is the pooled sample standard deviation, and nA and nB are the sample sizes.

B: To create a 98% confidence interval for the difference in the placebo and the new drug, we will use the formula:

CI = (xB - xA) ± tα/2,sP × √(1/nA + 1/nB)

where xA and xB are the sample means, sP is the pooled sample standard deviation, nA and nB are the sample sizes, and tα/2,sP is the t-value corresponding to a 98% confidence level with 18 degrees of freedom.

Using the values from Part A, we have:

CI = (18.3 - 9.7) ± 2.101 × 8.064 × √(1/10 + 1/10)

= 8.6 ± 5.103

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The correct question is:

Twenty middle-aged men with glucose readings between 90 milligrams per deciliter and 120 milligrams per deciliter of blood were selected randomly from the population of similar male patients at a large local hospital. Ten of the 20 men were assigned randomly to group A and received a placebo. The other 10 men were assigned to group B and received a new glucose drug. After two months, posttreatment glucose readings were taken for all 20 men and were compared with pretreatment readings. The reduction in glucose level (Pretreatment reading − Posttreatment reading) for each man in the study is shown here.

Group A (placebo) reduction (in milligrams per deciliter): 12, 8, 17, 7, 20, 2, 5, 9, 12, 6

Group B (glucose drug) reduction (in milligrams per deciliter): 29, 31, 13, 19, 21, 5, 24, 12, 8, 21

Part A: Do the data provide convincing evidence, at the α = 0.02 level, that the glucose drug is effective in producing a reduction in mean glucose level?

Part B: Create and interpret a 98% confidence interval for the difference in the placebo and the new drug.

Therefore , the solution of the given problem of equation comes out to be the ball is 2.25 inches tall after the fifth bounce.

In order to demonstrate consistency between two opposing statements, variable words are frequently used in sophisticated algorithms. Equations are academic phrases that are used to demonstrate the equality of different academic figures. Consider expression the details as y + 7 offers. In this case, elevating produces b + 7 when partnered with building y + 7.

Here,

a)

We are aware that the ball was dropped from a height of 6 feet, or 72 inches, and that it bounced twice, the first time for 36 inches and the second time for 18 inches. We can thus write:

=> a = 72 (the initial height)

=> B = (height of previous bounce minus height of each subsequent bounce) = 18/36 = 1/2

As a result, the following exponential equation can be used to predict the ball's height:

=> y = 72*(1/2)ˣ

b) We enter x = 5 into the equation we derived in part a) to determine the height of the ball on the fifth bounce:

=> y = 72*(1/2)⁵

=> y = 72*(1/32)

=> 2.25 inches for y.

As a result, the ball is 2.25 inches tall after the fifth bounce.

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The shaded area of the circle is around 65.44 square meters in size.

To find the area of the shaded region, subtract the area of sector FGH from the area of sector FEGH.

The area of sector FEGH is:

A1 = (1/2) r² θ₁

where r = radius of the larger circle and θ₁ = angle subtended by the arc EH.

Since EH = 30 m and the radius of the larger circle = 18 m (half of 10 + 8):

θ₁ = (EH arc length) / r = 30/18π radians

So,

A₁ = (1/2) (18)² (30/18π) = 270/π m²

The area of sector FGH is:

A₂ = (1/2) r² θ₂

where θ₂ = angle subtended by the arc GH.

Since GH is 8 m and the radius of the larger circle is 18 m:

θ2 = (GH arc length) / r = 8/18π radians

So,

A₂ = (1/2) (18)² (8/18π) = 64/π m²

Therefore, the area of the shaded region is:

A = A₁ - A₂ = (270/π) - (64/π) = 206/π ≈ 65.44 m²

Hence, the area of the shaded region is approximately 65.44 square meters.

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The explicit rule for the geometric sequence with a recursive rule of a1=4/9 and an=3an-1 is: an = (4/9) * (3)^(n-1)

To find the explicit rule for a geometric sequence, we use the formula:

an = a1 * r^(n-1)

where a1 is the first term, r is the common ratio, and n is the term number.

Given the recursive rule for this geometric sequence as a1=4/9 and an=3an-1, we can find the common ratio:

an = 3an-1

an/an-1 = 3/1

r = 3

Now we can use the formula to find the explicit rule:

an = a1 * r^(n-1)

an = (4/9) * (3)^(n-1)

Therefore, the explicit rule is: an = (4/9) * (3)^(n-1)

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Answer:

301.4 in^3

Step-by-step explanation:

Divide 12 by 2 to get the radius

12/2=6

Square 6 and multiply by 3.14 according to the area of a circle formula πr^2.

6^2=36 36*3.14=113.04

We now have the area of the base.

Multiply it by the height and divide by three to follow the formula for the volume of a cone, 1/3bh.

113.04*8=904.32. 904.32/3= 301.44, or 301.4 after rounding.

Answer:

I am sorry this is just so new for me like what even is an "imaginary" solution, i am in 6th grade wth

The simplification of the function is r(x)=10x - 95.

We are given that;

s(x) = -85 and a(x) = 10(x - 1)

Now,

To create a function that combines them, you can substitute these expressions into the formula above:

r(x) = s(x) + a(x) r(x) = (-85) + 10(x - 1)

You can simplify this function by distributing the 10 and combining the constants:

r(x) = -85 + 10x - 10 r(x) = 10x - 95

Therefore, the function will be r(x)=10x - 95.

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The estimated number of students per grade level in the state is between 66,700 and 112,800 students, with an estimated total of 580,450 students.

The anticipated number of students in line with grade level within the state is between:

(178.6 - 45.2) x 500 = 66,700 students

and

(178.6 + 45.2) x 500 = 112,800 students.

This is the range of values that we can be 95% confident includes the actual average number of students according to grade degree within the state, based on the sample of 500 high schools and the margin of mistakes of ±45.2.

To estimate the whole number of students according to grade level in the state, we can multiply the anticipated average with the aid of the total number of high schools:

3250 x 178.6 = 580,450 students

Consequently, the estimated number of students per grade level in the state is between 66,700 and 112,800 students, with an estimated total of 580,450 students.

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Answer:

a) 27. b) 27.3333 c) 40.75

The volume of the cylinder is 8,038.4 ft3

The height(H) is 10ft

The diameter is 32 ft

The formula for calculating the volume of a cylinder is

πr²h

Radius= 32/2

= 16

Volume= 3.14 × 16² × 10

= 3.14 × 256 × 10

= 8,038.4

Hence the volume of the cylinder is 8,038.4 ft3

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Answer:

Slope = 14.000/2.000 = 7.000 x-intercept = 3/7 = 0.42857 y-intercept = -3/1 = -3.00000

y intercept = -3

slope = 7/1

Note that the angle measurements are given as follows:

A = 145° (Sum of angles on a straight line.

B = 35° Opposite angles

C = 145° Opposite angels

D) = 35° supplementary angels.

a) Note that Sum of angles on a straight line=180° and are therefore supplementary

b) All opposite angle are equal, since ∠b is opposite to 35°, then ∠b = 35°

c) 145° is also opposite to ∠c

d) when two angles are supplementary, it means that they sum up to 180°

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Full Question:

Solve for the missing angle measurements for angles a, b, c, and d
See the attached imge

Answer:

Step-by-step explanation:

Given:

q(x) = -3x+4

r(x) = -4x

(r₀q)(4)   =>    r(q(4))     solve for q(4) first

q(4) = -3(4)+4       > substitute 4 in for x

q(4) = -12 +4

q(4) = -8           > now substitute this into r(q(4))

r(-8) = -4(-8)     > substitute -8 in for x

r(-8) = 32

(r₀q)(4) = 32

(q₀r)(4)   => q(r(4))    solve for r(4) first

r(4) = -4(4)        > substitute 4 for x

r(4) =-16            > substitute this into q(x)

q(-16) = -3(-16)+4

q(-16) = 48 +4

q(-16) = 52

(q₀r)(4) = 52