find what is x + 2½ = 10?

Given:

The table of data values.

To find:

What is the percentage of 10th grade students are in 2 clubs.

Solution:

From the given table it is clear that the number of 10th grade students are in 2 clubs is 12.

Total number of students in 10th grade is:

\(125+12+15=152\)

Now, the percentage of 10th grade students are in 2 clubs is:

\(\text{Required percentage}=\dfrac{\text{Number of 10th grade students are in 2 clubs}}{\text{Total number of students in 10th grade}}\times 100\)

\(\text{Required percentage}=\dfrac{12}{152}\times 100\)

\(\text{Required percentage}=\dfrac{1200}{152}\)

\(\text{Required percentage}\approx 7.89\%\)

Therefore, the percentage of 10th grade students are in 2 clubs is 7.89%.

Answer:

x³-25x²+3x-12

___________

x

Answer:

12.496

Step-by-step explanation:

Here, to calculate the sum of the first five terms of the series, we use the formula for the sum of terms in a geometric series.

Mathematically, that would be;

Sn = a(1-r^n)/1-r

where Sn = ?

a = 10

r = 1/5 = 0.2

n = 5

Substituting these values into the equation above, we have;

Sn = 10(1-(0.2)^5)/1-0.2

Sn = 10(0.99968)/0.8

Sn = 12.496

Answer:

Step-by-step explanation:

Integrating the inner integral with respect to x yields the required value of 32/3.

Given integral is: \($$\int_{0}^{2}\int_{1}^{3}x^2y dydx$$\)

Let's evaluate the integral by reversing the order of integration and integrating with respect to y first.

\($$= \int_{1}^{3}\int_{0}^{2}x^2y dxdy$$\)

Now we integrate the inner integral with respect to x.

\($$= \int_{1}^{3}\frac{1}{3}x^3y \bigg|_{x=0}^{x=2} dy$$\)

Substituting the limits, we get;

\($$= \int_{1}^{3}\frac{8}{3}y dy$$$$\)

\(= \frac{8}{3} \cdot \frac{y^2}{2} \bigg|_{1}^{3}$$$$\)

\(= \frac{8}{3} \cdot \frac{9}{2} - \frac{8}{3} \cdot \frac{1}{2}$$$$\)

\(= \frac{8}{3} \cdot 4$$$$\)

\(= \frac{32}{3}$$\)

Thus, the required value of the given integral is 32/3.

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Answer:

Step-by-step explanation:

1) x=14 y=40

2)a=10 b=37

3)u=62 v=59

4)s=9 t=14

5)u=66 v=38

hope this helps

X is a geometric random, its variable is constant and equal to p.

To show that if X is a geometric random variable, its failure rate is constant and equal to p, we will use the definition of the geometric distribution and the failure rate formula.

As the expression for z(x) does not depend on x, the failure rate is constant and equal to p/(1-p). However, we need to show that it is equal to p.
Recall that p is the probability of success. Since failure and success are complementary events, the probability of failure is q = 1-p.

Observe that the failure rate z(x) equals the probability of success p when divided by the probability of failure q. Since the failure rate for a geometric random variable only depends on the probability of success p and the probability of failure q, it is constant and equal to p.

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In order to find the price increase in percentage terms, we divide the price difference by the cost price.

The difference:

Rate:

If we solve for \(x\) in this equation, it will give us the price increase as a percentage.

The price increase should be \(110\)%.

Answer:

Z+12a−z+8+12−+8

Step-by-step explanation:

Z+8−6(z−2a)+5z

+8−6(−2)+5

Z+8−6(−2a+z)+5z

+8−6(−2+)+5

Z+8−6(−2a+z)+5z

+8−6(−2+)+5

Z+8+12a−z+8+12−

Answer:

Z + 8 + 12a − z

12a + 8

Step-by-step explanation:

question 1 x-y=-2

Slope:  1

y-intercept: (0,2)

X=0,Y=2

X=1 Y=3

question 2  2x-2y=-4

slope =1

y-intercept: (0,2)

X=0,Y=2

X=1 Y=3

Everything is the same

Answer:

B. 5

Step-by-step explanation:

_________________

Answer:

B)

Step-by-step explanation:

y = 5x - 22

y = mx + c

So the slope is 5

Answer:

It would look about like this

Step-by-step explanation:

On substituting the given values, da/dt = (11.7 cm/s) × √3a cm

= (11.7 cm/s) × √3 × 1.0 cm

= 20.25 cm/s. The diagonal of the cube is increasing at a rate of 20.25 cm/s. Answer: 20.25 words.

Given: Rate of increase in edge of a cube = 11.7 cm/s. The edge of a cube is increasing at a given rate. Now, we need to find out at what rate the diagonal of a cube is increasing.

Let a be the edge of the cube. Diagonal of the cube d = √(a² + a² + a²)

= √3aWe are given that the edge of the cube is increasing at a rate of 11.7 cm/s. So, da/dt = 11.7 cm/s. We need to find out the rate of change of the diagonal of a cube. Let's differentiate the above equation with respect to t,da/dt = (d/da) × √3a × da/d

t= √3 × da/dt × a

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The edge of the cube is increasing at a rate of 11.7 cm/s. So, da/dt = 11.7 cm/s. The rate at which the diagonal of a cube is increasing is approximately 33.1 cm/s.

On substituting the given values, da/dt = (11.7 cm/s) × √3a cm

= (11.7 cm/s) × √3 × 1.0 cm

= 20.25 cm/s.

The diagonal of the cube is increasing at a rate of 20.25 cm/s.

Rate of increase in edge of a cube = 11.7 cm/s. The edge of a cube is increasing at a given rate. Now, we need to find out at what rate the diagonal of a cube is increasing.

Let a be the edge of the cube. Diagonal of the cube

d = √(a² + a² + a²)

    = √3a

We are given that the edge of the cube is increasing at a rate of 11.7 cm/s. So,

da/dt = 11.7 cm/s.

We need to find out the rate of change of the diagonal of a cube. Let's differentiate the above equation with respect to t,

da/dt = (d/da) × √3a × da/d

t= √3 × da/dt × a

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Answer:

x² + 34x + 104 = 0

x² + 34x = -104

x² + 34x + ((1/2)(34))² = -104 + ((1/2)(34))²

x² + 34x + 17² = -104 + 17²

x² + 34x + 289 = 185

(x + 17)² = 185

x + 17 = +√185

x = -17 + √185

You can use either a numerical (integer or floating-point) variable or a Boolean variable to store the value of a logical expression.

Numerical Variable: You can use a numerical variable, such as an integer or floating-point variable, to store the result of a logical expression. In this case, the logical expression would be evaluated and assigned a numerical value, typically 0 or 1, representing false or true, respectively. For example, if you have a logical expression "x > 5", you can assign the result to a numerical variable like "result = (x > 5)", where the value of "result" would be 0 if the expression is false and 1 if it is true.

Boolean Variable: Alternatively, you can use a Boolean variable to directly store the truth value of a logical expression. A Boolean variable can only have two possible values: true or false. In this case, the logical expression would be evaluated and directly assigned to the Boolean variable. For example, if you have a logical expression "x > 5", you can assign the result to a Boolean variable like "isGreaterThanFive = (x > 5)", where "isGreaterThanFive" would be true if the expression is true and false if it is false.

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The perimeter of the figure is 4d+6c - 6

The distance around the outside of a 2D shape is known as its perimeter. Add together the lengths of all the sides to determine the perimeter.

We have been given a figure. We are asked to write an expression for the perimeter of our given figure.

We know that perimeter of a solid is sum of all of its side.

We can see that opposite side of our given figure are equal and our given figure is a rectangle.

We know that perimeter of rectangle is two times the length and width of rectangle, so the perimeter of our given figure would be:

2×(2d- 3+3c)

Use distributive property:

2 × 2d - 2 × 3 + 2 × 3 c

4d -6+6c

4d+6c-6

Therefore, the perimeter of our given figure is 4d+6c - 6

The complete question is : Write and simplify an expression for the perimeter of each figure

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Answer:

-3/4

Step-by-step explanation:

6x+8y=16

(subtract 6x from both sides)

8y=-6x+16

(divide everything by 8)

y=-6/8x+16

(simplify)

y= -3/4x+16

this is now in y=mx+b form. m=slope

Answer:

The solution is A. (4, -2)

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

Answer: well

Step-by-step explanation:

Answer:

52 degrees

Step-by-step explanation:

<JRK and <GRO are vertical angles so that means that they have equal angle measures.

The definite integral \(\int_{1}^{2}\left(x-\frac{1}{x}\right)^{2} dx\) evaluates to 1.500.

Step 1: Expand the integrand: \(\left(x-\frac{1}{x}\right)^{2} = x^{2} - 2x\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^{2} = x^{2} - 2 + \frac{1}{x^{2}}\).

Step 2: Integrate each term of the expanded integrand separately.

The integral of \(x^{2}\) with respect to \(x\) is \(\frac{x^{3}}{3}\).

The integral of \(-2\) with respect to \(x\) is \(-2x\).

The integral of \(\frac{1}{x^{2}}\) with respect to \(x\) is \(-\frac{1}{x}\).

Step 3: Evaluate the definite integral by substituting the upper limit (2) and lower limit (1) into the antiderivatives and subtracting the results.

Evaluating the definite integral, we have \(\int_{1}^{2}\left(x-\frac{1}{x}\right)^{2} dx = eft[frac{x^{3}}{3} - 2x - \frac{1}{x}\right]_{1}^{2} = \frac{8}{3} - 4 - frac{1}{2} - \left(\frac{1}{3} - 2 - 1\right) = \frac{4}{3} - \frac{1}{2} = \frac{5}{6} = 1.500\) (rounded to 3 decimal places).

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Answer:

estimated difference = 72

actual difference = 72.74

Step-by-step explanation:

Hello There!

First lets find the actual difference

Note: difference is subtraction

79.44-6.7=72.74

so the actual difference is 72.74

now lets find the estimated difference

79.44 rounded to the nearest whole is 79

6.7 rounded to the nearest whole is 7

79-7=72

so the estimated difference is 72

Answer:

x=5

y=5

Step-by-step explanation:

y-2=3

3x-2y=5

Let's solve for y in the first equation.

y-2=3

Add 2 to both sides,

y=5

Let's substitute y=5 into the other equation.

3x-2y=5

3x-2*5=5

3x-10=5

Add 10 to both sides,

3x=15

Divide by 5 to both sides,

x=5

Therefore, x and y both equal 5.

After the 10th iteration, the estimated value of the root obtained by the bisection method is approximately 0.517. Option B

To solve the equation cos(x) - x * e = 0 in the interval [0, 1] using the bisection method, we start by checking the function values at the endpoints of the interval.

For x = 0:

cos(0) - 0 * e = 1 - 0 = 1

For x = 1:

cos(1) - 1 * e ≈ 0.54 - 2.72 ≈ -2.18

Since the function values at the endpoints have opposite signs, we can apply the bisection method to find the root within the interval.

The bisection method involves repeatedly dividing the interval in half and checking the function value at the midpoint until a sufficiently accurate approximation is obtained. In this case, we will perform 10 iterations.

After the 10th iteration, the estimated value of the root obtained by the bisection method is approximately 0.517.

Therefore, the correct answer is OB) 0.517.

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Answer:

Prime factorization of 28 and 35 is (2 × 2 × 7) and (5 × 7) respectively. As visible, 28 and 35 have only one common prime factor i.e. 7.

Answer:

Unknown, Solve for x or y?

Step-by-step explanation:

You need to elaborate on your answer... sorry.

In the meantime, here is Microsoft Math Solver, With your equations:

https://mathsolver.microsoft.com/en/solve-problem/x-y%20%3D%20%203

You can use this tool next time :)

Have a great day,

Nate

Answer:

Option C is your answer

Step-by-step explanation:

21,22,  23,  26,29,  32,  34,38,  40,  42,43

Yes, Q1 is 23, median is 32, Q3 is 40

So the box should start on 23, end on 40, and have its inner line over 32.

In your picture, that's option C

Answer:its c

Step-by-step explanation:i took the test

Answer:

The grade of this sixth and (final) test would be 91.

Step-by-step explanation:

This is an arithmetic series since each time the data is increasing by the common difference of 4, and we already know a₁ which is the first term in the sequence is 71, therefore we can say:

a₁ = 71

r = 4

a₆ = the sixth term that needs to be found out

n = 6

Formula of an arithemetic series:

aₙ = a₁ + r(n - 1)

a₆ = 71 + 4(6 - 1)

a₆ = 71 + 4(5)

a₆ = 71 + 20

a₆ = 91

Hence, the grade of this sixth and (final) test would be 91.

Hope this helps!

The class that lost the most pencils overall based on the data displayed is D. Mr. Johnson's class; it has a wide spread in the data

The answer is Mr. Johnson's class. The median is the middle value in a set of data. In Mr. Johnson's class, the median is 11 pencils. This means that half of the students in his class lost 11 or fewer pencils, and half of the students lost 11 or more pencils.

In Mr. Simpson's class, the median is 14.5 pencils. This means that half of the students in his class lost 14.5 or fewer pencils, and half of the students lost 14.5 or more pencils.

Since the median for Mr. Johnson's class is lower than the median for Mr. Simpson's class, we can conclude that Mr. Johnson's class lost more pencils overall.

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