Solve for p.
r-P/7= S

Step-by-step explanation:

10 pieces, each 2/5 m long are

10 × 2/5 = 20/5 = 4 m

that leaves 8 - 4 = 4 m of ribbon.

now, from these 4 m, pieces of 3/4 m are cut.

how many ?

as many as times 3/4 fit into 4 :

4 / 3/4 = 4/1 / 3/4 = 4×4 / (1×3) = 16/3 = 5 1/3.

that means, she got max. 5 pieces of 3/4 m. and she has 1/3 m ribbon left.

The vertex form of a quadratic equation can be viewed as f(x) = a(x - p)^2 + q, where (p,q) corresponds to the vertex.

For instance, taking into account the equation f(x) = x² yields the result f(x) = 1(x - 0)^2 + 0, and thus the vertex is (0,0).

We complete the square in order to transform the quadratic into its vertex form as follows:

f(x) = x² - 8x - 17

f(x) = (x² - 8x + 16) - 16 - 17 (through addition and subtraction of 16 for completion of the square)

f(x) = (x - 4)² - 33

Therefore, the equation is expressed as f(x) = a(x - p)² + q, in which a = 1, p = 4, and q = -33.

It should be noted that to uncover the y-intercept of a function, we employ the strategy of setting x = 0, followed by calculating f(0):

f(x) = 4x² - 5x + 5

f(0) = 4(0)² - 5(0) + 5

f(0) = 5

Consequently, the y-intercept is revealed to be (0,5). Answer: A (0,5).

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Answer:

The sample is not random because he only poll the person that bought a hot lunch on a particular day and not random people so the sample is not random.

The distribution of the data shown on the dot plot is Skewed.

We have to given that;

The data shown on the dot plot.

Now, We know that;

A graph is symmetric if the left and right side of the graph are mirror images of each other. A graph is skewed if it has more data on one side rather than the other.

Here, it has more data on one side rather than the other.

Hence, the distribution of the data shown on the dot plot is Skewed.

Thus, the distribution of the data shown on the dot plot is Skewed.

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The solution for the given linear-pair of equations is (1,6) , and it has one number of solutions. The intercepts on x & y axis to plot the graph

i)3x - 3y + 15 = 0 : Point-1: (0,5) & Point-2:(-5,0)

ii)4x + 2y = 16 :Point-1:(0,8) & Point-2: (4,0)

What is linear-pair?

It is a system of continuous lines that looks like straight lines. The linear equation with two variables has a maximum degree of one. Ax+by+c=0 is the general form of a two-variable linear equation. A consistent pair of linear equations is a pair of linear equations in two variables that have a solution. To a pair of equivalent linear equations, there are countless possible common solutions. Such a pair is a dependent set of two-variable linear equations.

Given linear-pair :

3x - 3y + 15 = 0

3x - 3y = -15      ---- eqn-(i)

4x + 2y = 16      -----eqn-(ii)

Using elimination method: to equate the coefficient of x in both equations,

Multiplying equation -(i) by 4

(3x - 3y = -15 ) ⇒4(3x - 3y) = 4(-15)

                       ⇒12x -12y = -60 ------eqn(iii)

Multiplying equation -(ii) by 3

4x + 2y = 16 ⇒ 3(4x + 2y = 16)

                   ⇒3(4x + 2y ) = 3(16)

                   ⇒12x + 6y = 48 ------eqn(iv)

Subtracting eqn-iv from eqn-iii

12x -12y - (12x + 6y) = -60 - 48

12x - 12y - 12x - 6y = -108

-12y - 6y = -108

-18y = -108

18y = 108

y = 6

Substituting y=6 in eqn-(i) 3x - 3y = -15

3x - 3(6) = -15

3x - 18 = -15

3x = -15 + 18

3x = 3

x = 1

The solution for the given pair of equations is (1,6)

Points to plot graph:

3x - 3y = -15  ---- eqn-(i)

Taking x=0,

3(0) - 3y = -15

3y=15

y=5

Point-1: (0,5)

Taking y=0;

3x - 3(0) = -15

3x = -15

x = -5

Point-2:(-5,0)

4x + 2y = 16      -----eqn-(ii)

Taking x=0

4(0) +2y=16

2y=16

y=8

Point-1:(0,8)

Taking y=0

4x + 2(0) = 16

4x = 16

x = 4

Point-2: (4,0)

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refer to attachment for graph.

The lower endpoint of the 95% confidence interval is approximately 0.5058, meaning that we are 95% confident that the true proportion of voters in the school district who support increasing property taxes to maintain athletic programs is at least 50.58%.

To construct a 95% confidence interval for the proportion of voters who support increasing property taxes to maintain athletic programs, we can use the formula:

CI = p ± z*(√(p*(1-p)/n))

Where CI is the confidence interval, p is the sample proportion (115/200 = 0.575), z is the z-score for the desired confidence level (1.96 for 95% confidence), and n is the sample size (200).

Plugging in the values, we get:

CI = 0.575 ± 1.96*(√(0.575*(1-0.575)/200))
CI = 0.575 ± 0.079

So the 95% confidence interval for the proportion of voters who support increasing property taxes to maintain athletic programs is (0.496, 0.654). The lower endpoint of the interval is 0.496.

To construct a 95% confidence interval for the proportion of voters who support increasing property taxes to maintain athletic programs, follow these steps:

1. Calculate the sample proportion (p): Divide the number of voters who said yes (115) by the total number of voters in the sample (200).
  p = 115/200 = 0.575

2. Determine the sample size (n): In this case, n = 200.

3. Calculate the standard error (SE): SE = sqrt[(p(1-p))/n]
  SE = sqrt[(0.575(1-0.575))/200] ≈ 0.0353

4. Find the critical value (z) for a 95% confidence interval: z = 1.96 (You can find this value in a z-table or by using an online calculator.)

5. Calculate the margin of error (ME): ME = z * SE
  ME = 1.96 * 0.0353 = 0.0692

6. Determine the lower endpoint of the 95% confidence interval: Lower Endpoint = p - ME
  Lower Endpoint = 0.575 - 0.0692 ≈ 0.5058

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If one dozen avocados the same size weigh 7. 2 lb then the weight of one avocado is 0.6lb.

Weight is a measurement of the force of gravity acting on an item. It is the result of the object's mass and acceleration brought on by gravity. Usually, weight is expressed in units like pounds (lb) or kilograms (kg). Weight may be defined more simply as the amount of pressure an object applies to a scale when it is placed there. The force produced by a person's body on a scale as a result of gravity, for instance, is equal to 150 lb if they weigh 150 lb.

One dozen avocados equals 12 avocados. If one dozen avocados weigh 7.2 lb, then we can find the weight of one avocado by dividing the total weight (7.2 lb) by the number of avocados (12):

Weight of one avocado = Total weight / Number of avocados

Weight of one avocado = 7.2 lb / 12

Weight of one avocado = 0.6 lb

Therefore, the weight of one avocado is 0.6lb.

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Because area calculations always involve multiplying two lengths somewhere in the formula. The lengths must be in the same units and that gives length times length or length^2 units.

[tex]\textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ V=384 \end{cases}\implies 384=\cfrac{4\pi r^3}{3}\implies 384(3)=4\pi r^3 \\\\\\ \cfrac{384(3)}{4\pi }=r^3\implies \sqrt[3]{\cfrac{384(3)}{4\pi }}=r\implies 4.5\approx r[/tex]

Answer:

[tex] \frac{4}{3} \pi {r}^{3} = 384[/tex]

[tex]r = 4.5 [/tex]

So the radius of this sphere is approximately 4.5 feet.

Answer: 5 meters

Step-by-step explanation:

volume of a rectangular prism: V=whl

90=6*h*3

h=90/(6*3)

h=90/18

h=5 meters

The given expression is needed to be written in the the form a^n.

The required expression is a ^ -5/4

We have,

The given expression is

1/ a^5/4

We know,

the identity that a negative exponent means the reciprocal.

1/ a^b = a^-b

So, the given expression is

1/ a^5/4

=a ^-5/4

The required expression is a ^-5/4

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We fail to reject the null hypothesis and conclude that there is enough evidence to support the claim that the mean height of top-rated roller coasters is 160 feet.

The sample mean is 164.6111, which is the average height of the 36 roller coasters in the sample.

The test statistic is 0.3864, which is calculated as (sample mean hypothesized mean) / standard error.

The p-value is 0.6965, which is the probability of getting a test statistic as extreme as 0.3864, assuming that the hypothesized mean is true.

Since the p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis and conclude that there is enough evidence to support the claim that the mean height of top-rated roller coasters is 160 feet.

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The interquartile range (IQR) of the data is, 1.5, 2.5, 1.5 kilograms under the condition that a left whisker extends from 4. 5 to 6. The box extends from 6 to 8. 5 and is divided into 2 parts by a vertical line segment at 7.

Interquartile range (IQR) is the measurement of statistical dispersion based on splitting data and information, into four parts called quartiles.
To find the interquartile range (IQR), we have to evaluate the difference between the upper quartile (Q3) and lower quartile (Q1)

For the following case there are 3 cases, for the first case left whisker extends from 4. 5 to 6, for the second case it is given that the box extends from 6 to 8.5 and is divided into two parts by a vertical line segment at 7.  and for the third case right whisker extends from 8. 5 to 10.
Therefore,
Q1 = 4.5  
Q2 = 6
Q3 = 8.5

Q4 = 10
Then the evaluated  IQR for first case is
Q2 - Q1
= 6 - 4.5
= 1.5 kilograms

for second case is

Q3 - Q2

= 8.5 - 6

= 2.5  kilograms

finally for the third case

Q4 - Q3

= 10 - 8.5

= 1.5 kilograms

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The complete question is

Here's a box plot that summarizes the masses of Suika's watermelons. A horizontal boxplot is plotted along a horizontal axis marked from 2 to 10, in increments of 0. 5, labeled Mass (kilograms). A left whisker extends from 4. 5 to 6. The box extends from 6 to 8. 5 and is divided into 2 parts by a vertical line segment at 7. The right whisker extends from 8. 5 to 10. All values estimated. Find the interquartile range (IQR) of the data

The scatter plot is solved and

According to the data, a bowler will see an increase in their score since the slope of the line of best fit is positive.

By week 4, they should see a mean score of approximately 214 (rounded to the nearest whole number).

Overall, they should see an increase in their high bowling scores from week 1 to week 5 of approximately 35 (rounded to the nearest whole number) pins overall.

Given data ,

A bowling team tracks the high scores for the five members to see how their game progresses throughout a 5-week time period.

Let the scatter plot be represented as A

Now , the value of A is

The x-axis is labeled Weeks and has a scale ranging from 1 to 6. The y-axis is labeled Score and has a scale ranging from 170 to 260 in 10 unit increments. There are points at (1, 180), (1, 190), (1, 205), (1, 215), (1, 220), (2, 185), (2, 195), (2, 210), (2, 220), (2, 225), (3, 185), (3, 195), (3,210), (3, 222), (3, 228), (4, 188), (4, 200), (4, 212), (4, 230), (4, 235), (5, 195), (5, 210), (5, 220), (5, 230), and (5, 245). The line is drawn through points (0.5, 200), (3, 210), and (5.5, 220).

So , a bowler will see an increase in their score since the slope of the line of best fit is positive.

And , by week 4, they should see a mean score of 214

Now , an increase in their high bowling scores from week 1 to week 5 of approximately 35

Hence , the scatter plot is solved

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The dimensions of the rectangle are 8cm by 16cm

Remember that for a rectangle of length L and width W, the perimeter is.

P = 2*(L + W)

Here we know that:

L = 2*W

And that the perimeter is 48cm, so:

48cm = 2*(L + W)

So we have a system of equations:

L = 2*W

48cm = 2*(L + W)

Replacing the first equation in the second one, we will get:

48 cm = 2*(2W + W)

48cm = 6W

48cm/6 = W

8cm = W

And the length is:

L = 2W = 2*8cm = 16cm

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Answer:

The second answer: histogram with the title favorite sport and the x axis labeled sport and the y axis labeled number of students, with the first bar labeled baseball going to a value of 17, the second bar labeled basketball going to a value of 12, the third bar labeled tennis going to a value of 27, and the fourth bar labeled soccer going to a value of 44

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The area of the rhombus in the image given which has diagonal lengths of 6 cm and 8 cm respectively, is calculated as: 27 cm².

A rhombus is a type of a quadrilateral that has four equal sides with diagonals that bisect each other. To find the area of a rhombus, apply the formula below:

Area of a rhombus = ½ * (product of the lengths of the diagonals)

Thus, from the image given to us, the lengths of the two diagonals of the rhombus are 6 cm and 9 cm respectively. Therefore, we have:

Area of the rhombus = ½ * (6) * (9)

= 1/2 * 54

Area = 27 cm²

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The axis of symmetry is x = 5 and five points are drawn in the graph.

The given equation is

y = x² - 10x + 24

Write the equation as

   y = x² - 2(5)x + 2² - 2²+ 24

⇒ y = (x-5)² - 1

Since we know that for graph

y = a(x-a)² + h vertex is (h, k)

so vertex of graph y be (5 , -1)

Now after plotting of graph we have

Axis of symmetry be x = 5

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The solution is: matching the left with its equivalent expression on the right, is: a - 3, b - 4 , c - 1 , d - 5, e - 2.

Given that,

a. ✓4x²y^4.     1. 2x✓y

b. ✓8x²y.        2. 2y✓2x

c. ✓4x²y.         3. 2xy²

d. ✓16xy².       4. 2x✓2y

e. ✓8xy².        5. 4y✓x

so, multiplying we get,

a. ✓4x²y^4 = ✓4 × ✓x² × ✓y^4

= 2 × x × y²

= 2xy²

Therefore, ✓4x²y^4 = 2xy²

b. ✓8x²y = ✓8 × ✓x² × ✓y

= ✓4 × ✓2 × ✓x² × ✓y

= 2 × ✓2 × x × ✓y

= 2x✓2y

Therefore, ✓8x²y = 2x✓2y

c. ✓4x²y = ✓4 × ✓x² × ✓y

= 2 × x × ✓y

= 2x✓y

Therefore, ✓4x²y = 2x✓y

d. ✓16xy² = ✓16 × ✓x × ✓y²

= 4 × ✓x × y

= 4y✓x

Therefore, ✓16xy² = 4y✓x

e. ✓8xy² = ✓8 × ✓x × ✓y²

= ✓4 × ✓2 × ✓x × ✓y²

= 2 × ✓2 × ✓x × y

= 2y✓2x

Therefore, ✓8xy² = 2y✓2x

Hence, The solution is: matching the left with its equivalent expression on the right, is: a - 3, b - 4 , c - 1 , d - 5, e - 2.

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complete question:

Match each expression on the left with an equivalent expression on the right

a. ✓4x²y^4.     1. 2x✓y

b. ✓8x²y.        2. 2y✓2x

c. ✓4x²y.         3. 2xy²

d. ✓16xy².       4. 2x✓2y

e. ✓8xy².        5. 4y✓x

Therefore, the state government should consider investing in new affordable housing initiatives.

a) To compute the 90% confidence interval, we can use the formula:

CI = p ± z*(√(p(1-p))/n)

where p is the sample proportion, z is the z-score corresponding to a 90% confidence level (which is 1.645), and n is the sample size.

Plugging in the values, we get:

CI = 0.517 ± 1.645*(√((0.517)(1-0.517))/4527)

CI = 0.505 to 0.529 (rounded to three decimal places)

b) The 90% confidence interval tells us that we are 90% confident that the true proportion of Californians who considered moving out of the state is between 0.505 and 0.529.

c) Since the confidence interval does not include the value of 0.5, which is the null hypothesis value, we can conclude that there is evidence to suggest that more than 50% of Californians would consider moving out of state due to high cost of housing.

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The value of r not equal to the radius of the circle, AB is not tangent to the circle.

A tangent to a circle is a straight line that precisely crosses the circle at the location where the tangency occurs. The tangent line must be perpendicular to the circle's radius in order to be at the point of tangency.

The circle's radius and the tangent line are related because at the point of tangency, the tangent line is always perpendicular to the radius. This indicates that at the point of tangency, the radius and tangent line make a right angle. In addition, the radius of the circle is equal to the length of the segment connecting the point of tangency and the centre of the circle. Tangent to circle difficulties can be resolved using these relationships.

In the given figure for AB to be the tangent we must determine whether the radius is perpendicular and has the same value of 9.

For the given triangle assume angle A is at 90 degree.

Thus,

12² + r² = 14²

144 + r² = 196

r² = 196 - 144

r = √52 = 7.2

Here, the given radius 9 is not equal to the calculated radius.

Hence, the value of r not equal to the radius of the circle, AB is not tangent to the circle.

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c =

a

sin(α)

= 51.13963

b = √c2 - a2

= √51.1396257719972 - 82

= √2551.2613240999

= 50.51001

∠β = 90° - ∠α

= 81° = 1.41372 rad = 9/20π

area =

a × b

2

=

8 × 50.5100121174

2

= 202.04005

perimeter = a + b + c

= 8 + 50.5100121174 + 51.139625771997

= 109.64964

inradius =

a × b

a + b + c

=

404.0800969392

109.6496378894

= 3.68519

circumradius =

c

2

=

51.139625771997

2

= 25.56981

Length of the aquarium is 11.25 ft

We have,

Volume of the aquarium = 625 ft³

given by

Volume = length × width × height.

Width = 16 2/3 ft

          = 50/3

and Height = 3 1/3 ft

                   = 10/3 ft

Length = Volume/Width × Height

           = 625/ 50/3 × 10/3

           = 625/ 500/9

           = 11.25 ft

Hence, Length of the aquarium is 11.25 ft

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The transformed parent function can be written as follows:

g(x) = -k*(x + N)

Here we start with the parent function:

f(x) = x

First, a horizontal translation of N units is written as:

g(x) = f(x + N)

A reflection over the y-axis just adds a negative sign to the function, then we get:

g(x) = -f(x + N)

And a vertical dilation of scale factor k is written as:

g(x) = -k*f(x + N)

replacing f(x) we get:

g(x) = -k*(x + N)

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The value of h in the right triangle is 3 units.

A right angle triangle is a triangle that has one of its angles as 90 degrees. The sum of angles in a triangle is 180 degrees.

Therefore, let's find h in the right triangle as follows:

Let's find the angle.

cos ∅ = adjacent / hypotenuse

cos ∅ = 6 / 6.5

∅ = cos⁻¹ 0.92307692307

∅ = 22.6208962498

Therefore,

sin 22.62° = h / 7.8

cross multiply

h = 7.8 × sin 22.62°

h = 7.8 × 0.3846175604

h = 3.00001697112

h ≈ 3 units

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Answer:

there is brainly free and other options if you do not want to watch ads. but it is free if you do not want to upgrade.

Step-by-step explanation:

Answer:

yes

Step-by-step explanation:

Answer:

4 students for the first

Step-by-step explanation:

i will finish answering a little bit later

The drawing and labelling of each figure are added as an attachment

Point E

By definition, a point is the location in any plane and is represented by a dot

Line segment BC

By definition, a line segment is a line that is bounded by two boundaries

Line VW

By definition, a line is a straight line that extends indefinitely

Angle DFG

This is formed by  the intersection of two lines or rays

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There are 36 possible combinations of the four models and nine colors to choose from.

To find the total number of combinations possible, we can use the formula: n(A ⨯ B) = n(A) x n(B), where n(A) is the number of models and n(B) is the number of colors to choose from.

In this case, there are four models and nine colors to choose from for each model. Therefore, the number of combinations possible would be:

n(A ⨯ B) = 4 x 9 = 36

So there are 36 possible combinations. This can also be calculated by using the formula n(A') = n(S) - n(A), where n(A') is the complement of the set A, n(S) is the total number of possible options, and n(A) is the number of options we are not interested in.

In this case, the complement of the set A (which represents all possible combinations) would be the empty set, since we are interested in all possible combinations. Therefore, n(A') = 0, and n(S) = 36.

We could also use the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B), where n(A ∪ B) represents the total number of options in both set A and set B, n(A) is the number of options in set A, n(B) is the number of options in set B, and n(A ∩ B) is the number of options that are common to both set A and set B.

In this case, we could consider set A to be the four models, and set B to be the nine colors for each model. The intersection of set A and set B would be the combinations where each model is paired with a specific color. Therefore, n(A ∩ B) = 4 x 9 = 36.

Using the formula, we can find the total number of options:

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

n(A ∪ B) = 4 + 9 x 4 - 36

n(A ∪ B) = 36

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The sunfish adds around 2% to its present mass each day, which is the daily percent change in mass.

Based on the provided model, the daily percent change in the mass of the ocean sunfish is an addition of around 2% to its mass. By calculating the derivative of the function M(t) with respect to t, the following may be calculated:

[tex]M'(t) = \frac{ln(1.35)}6 * (1.35)^t[/tex]

This indicates that the constant factor is ln(1.35)/6, or around 0.02, and therefore the rate of change of the sunfish's mass with respect to time is proportionate to the sunfish's present mass.

As a result, the sunfish's daily percent change in mass is around a 2% increase over its present mass.

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