draw a diagram and explain how you can use it to find 3 ÷ 1/5​

The vertex form of the quadratic equations in standard form are, respectively:

Case 9: y = 2 · (x + 2)² - 12

Case 10: y = - (1 / 2) · (x + 3 / 4)² + 33 / 32

Case 11: y = 3 · (x - 4 / 3)² - 16 / 3

Case 12: y = - 3 · (x - 3)²

Case 13: y = (x - 4)² + 3

Case 14: y = (x - 1)² - 7

Case 15: y = (x + 3 / 2)² - 9 / 4

Case 16: 2 · (x + 1 / 4)² - 1 / 8

Case 17: y = 2 · (x - 3)² - 7

Case 18: y = - 2 · (x + 1)² + 10

In this problem we find ten cases of quadratic equation in standard form, whose vertex form can be found by a combination of algebra properties known as completing the square. Completing the square consists in simplifying a part of the quadratic equation into a power of a binomial.

The two forms are introduced below:

Standard form

y = a · x² + b · x + c

Where a, b, c are real coefficients.

Vertex form

y - k = C · (x - h)²

Where:

Now we proceed to determine the vertex form of each quadratic equation:

Case 9

y = 2 · x² + 4 · x - 4

y = 2 · (x² + 2 · x - 2)

y = 2 · (x² + 2 · x + 4) - 12

y = 2 · (x + 2)² - 12

Case 10

y = - (1 / 2) · x² - 3 · x + 3

y = - (1 / 2) · [x² + (3 / 2) · x - 3 / 2]

y = - (1 / 2) · [x² + (3 / 2) · x + 9 / 16] + (1 / 2) · (33 / 16)

y = - (1 / 2) · (x + 3 / 4)² + 33 / 32

Case 11

y = 3 · x² - 8 · x

y = 3 · [x² - (8 / 3) · x]

y = 3 · [x² - (8 / 3) · x + 16 / 9] - 3 · (16 / 9)

y = 3 · (x - 4 / 3)² - 16 / 3

Case 12

y = - 3 · x² + 18 · x - 27

y = - 3 · (x² - 6 · x + 9)

y = - 3 · (x - 3)²

Case 13

y = x² - 8 · x + 19

y = (x² - 8 · x + 16) + 3

y = (x - 4)² + 3

Case 14

y = x² - 2 · x - 6

y = (x² - 2 · x + 1) - 7

y = (x - 1)² - 7

Case 15

y = x² + 3 · x

y = (x² + 3 · x + 9 / 4) - 9 / 4

y = (x + 3 / 2)² - 9 / 4

Case 16

y = 2 · x² + x

y = 2 · [x² + (1 / 2) · x]

y = 2 · [x² + (1 / 2) · x + 1 / 16] - 2 · (1 / 16)

y = 2 · (x + 1 / 4)² - 1 / 8

Case 17

y = 2 · x² - 12 · x + 11

y = 2 · (x² - 6 · x + 9) - 2 · (7 / 2)

y = 2 · (x - 3)² - 7

Case 18

y = - 2 · x² - 4 · x + 8

y = - 2 · (x² + 2 · x - 4)

y = - 2 · (x² + 2 · x + 1) + 2 · 5

y = - 2 · (x + 1)² + 10

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Answer: To solve this problem, we can use the binomial distribution formula. Let X be the number of blue jelly beans in a bag of 200 jelly beans. Then X follows a binomial distribution with parameters n = 200 and p = 0.15, where p is the probability of drawing a blue jelly bean.

The probability of getting more than 20% blue jelly beans in a bag can be calculated as:

P(X > 0.2*200) = P(X > 40)

We can use the normal approximation to the binomial distribution, since n is large (200) and p is not too close to 0 or 1. Using the mean and variance of the binomial distribution, we can calculate the corresponding mean and standard deviation of the normal distribution as follows:

μ = np = 200 * 0.15 = 30

σ = sqrt(np(1-p)) = sqrt(200 * 0.15 * (1-0.15)) = 4.07

Then, we can standardize the random variable X as:

Z = (X - μ) / σ

So, we have:

P(X > 40) = P((X - μ) / σ > (40 - μ) / σ)

= P(Z > (40 - 30) / 4.07)

= P(Z > 2.46)

Using a standard normal distribution table or a calculator, we find that P(Z > 2.46) is approximately 0.007. Therefore, the probability that a bag will contain more than 20% blue jelly beans is about 0.007 or 0.7%.

Step-by-step explanation:

a. The random variable X represents the number of people over 50 who ran and died in the same eight-year period.

b. The appropriate distribution for this problem is the binomial distribution, which describes the number of successes.

c. The 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight-year period is (0.0206, 0.1149).

d. The true population proportion lies within this interval, nor does it mean that the observed proportion (0.0677) has a 97% chance of being correct.

a. The random variable X represents the number of people over 50 who ran and died in the same eight-year period. The population proportion of over-50s who ran and passed away within the same eight-year period is represented by the random variable P'.

b. The appropriate distribution for this problem is the binomial distribution, which describes the number of successes (in this case, people who ran and died) in a fixed number of independent trials (in this case, the population of people over 50 during the eight-year period), where the probability of success is constant for each trial.

c. To construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight-year period, we can use the following formula:

\(\hat{p} \pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

where \hat{p} is the sample proportion, n is the sample size, and \(z_{\alpha/2}\) is the z-score associated with the desired level of confidence (97%, in this case).

Given that 1.5% of the 451 members of the 50-Plus Fitness Association died during the eight-year period, we can estimate the sample proportion as:

\(\hat{p} = \frac{X}{n} = \frac{0.015 \times 451}{1} = 6.765 \approx 0.0677\)

where X = \(0.015 \times 451 \approx 7\) is the observed number of people over 50 who ran and died in the same eight-year period.

Substituting the given values into the formula, we have:

\(0.0677 \pm 2.17\sqrt{\frac{0.0677(1-0.0677)}{451}} \approx (0.0206, 0.1149)\)

Therefore, the 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight-year period is (0.0206, 0.1149).

ii. The graph of the confidence interval would be a horizontal line segment between the two endpoints (0.0206, 0.1149) on the y-axis, with a vertical line at the midpoint (0.0677) on the x-axis.

iii. The error bound is:

\(z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \approx 0.0471\)

d. A "97% confidence interval" means that if we were to repeat this study many times, using different samples of the same size from the same population, we would expect the true population proportion of people over 50 who ran and died in the same eight-year period to be captured by the confidence interval in 97% of those repetitions. In other words, we are 97% confident that the true population proportion falls within the interval (0.0206, 0.1149). It does not mean that there is a 97% chance that the true population proportion lies within this interval, nor does it mean that the observed proportion (0.0677) has a 97% chance of being correct.

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if c is a irrational number that means a or b is also an irrational number or both of them are irrational numbers but they are different then each other

i hope that will help

There are two possible answers:

You can only pick one of those two cases. In short, exactly one of the 'a' or b is irrational and the other is rational. Both 'a' and b cannot be rational together, and they both cannot be irrational together.

================================================

Explanation:

For a moment, let's say that 'a' and b are both rational numbers. This means they would be ratios of integers. So,

a = p/q

b = r/s

where p,q,r,s are integers. The denominators q & s cannot be zero.

Multiplying those rational numbers 'a' and b leads to

a*b = (p/q)*(r/s) = (p*r)/(q*s)

We get a result in the form of integer/integer. Recall that multiplying any two integers leads to some other integer. Since (p*r)/(q*s) is a ratio of integers, it is rational and that makes a*b = c to be rational. However, your teacher said that c is irrational which is the opposite of rational.

In short, if both 'a' and b are rational, then a*b = c is rational. This means we know for certain that both 'a' and b cannot be both rational for c to be irrational.

------------------------

Next, we'll consider 'a' and b to both be irrational. Is it possible to have a*b = c to be irrational as well? It depends.

If we have the following

a = sqrt(3)

b = sqrt(12)

then,

a*b = sqrt(3)*sqrt(12) = sqrt(3*12) = sqrt(36) = 6

Showing that a*b is rational because 6 = 6/1 is a ratio of integers. This is a bit strange considering how 'a' and b themselves were irrational, but they led to a rational product.

This doesn't always happen since something like

sqrt(3)*sqrt(5) = sqrt(3*5) = sqrt(15)

is irrational. So if both 'a' and b are irrational together, then a*b could be rational or it could be irrational. We would need more information about the values of 'a' and b.

Therefore, we know that we can rule out the case that 'a' and b are irrational together.

------------------------

Lastly, we'll consider the case when one of 'a' or b is rational while the other is irrational. The order doesn't matter. So we'll make 'a' to be irrational and b to be rational.

Since b is rational, this means b = p/q for some integers p,q where q is nonzero. The value of 'a' cannot be written as a ratio like this because it is irrational.

Multiplying any nonzero rational number with an irrational one always leads to an irrational result. If a*b was rational, then we could say

a*b = r/s

a*(p/q) = r/s

a = (r/s)*(q/p)

a = (r*q)/(s*p)

Showing that 'a' is rational, but this contradicts the fact we made 'a' to be irrational at the top of this section. Therefore, a*b is always irrational if exactly one of 'a' or b is irrational and the other is rational.

In other words, if,

then a*b = c is irrational

Answer:

C

Step-by-step explanation:

Bc x 100 = percent conversion to decimal

Answer:

c

Step-by-step explanation:

Answer:

The probability that an individual has this disease given that the test indicates its presence is 0.1

Step-by-step explanation:

Let A be the event which indicates Having disease

Let B be the event in which test indicates that disease presence

Now we are given that 10% of the individuals in this age group suffer from this form of arthritis

So, P(A)=0.1

We are also given that the proposed test was given to individuals with confirmed arthritics disease, and a correct test result was obtained in 85% of the cases.

P(B|A)=0.85

Now we are given that. When the test was administered to individuals of the same age group who were known to be free of the disease, 4% were reported to have the disease

So,P(B|A')=0.04

We are supposed to find  the probability that an individual has this disease given that the test indicates its presence i.e. P(A|B)

So we will use Bayes theorem

\(P(A|B)=\frac{P(B|A) P(A)}{\sum P(B|A) P(A)}\)

\(P(A|B)=\frac{0.85 \times 0.1}{0.85 \times 0.1+0.85 \times 0.9}=0.1\)

Hence the probability that an individual has this disease given that the test indicates its presence is 0.1

Answer:

The correct option is (a).

Step-by-step explanation:

The Pearson's correlation coefficient is a statistical degree that computes the strength of the linear relationship amid the relative movements of the two variables (i.e. dependent and independent).It ranges from -1 to +1.

Positive correlation is an association amid two variables in which both variables change in the same direction.  A positive correlation occurs when one variable declines as the other variable declines, or one variable escalates while the other escalates.

Negative correlation is a relationship amid two variables in which one variable rises as the other falls, and vice versa.

In this case, it is provided that the Pearson correlation coefficient is, r = +0.88.

A correlation coefficient between ±0.70 to ±1.00 are considered as strong positive correlation.

The scatter plot for correlation coefficient between +0.70 to +1.00 shows:

Thus, the correct option is (a).

\(\huge\text{Hey there!}\)

\(\large\text{Simplify the expression: 2c+ 2 + 5c}\)

\(\large\text{2c+ 2 + 5c}\)

\(\large\text{COMBINE the LIKE TERMS}\)

\(\large\text{\bf 2c + 5c}\large\text{ + \underline{2}}\)

\(\large\text{\bf 2c + 5c = \huge 7c}\)

\(\large\text{\underline{2} = \huge \underline{2}}\)

\(\large\text{= \bf 7c }\large\text{+ \underline{2}}\)

\(\boxed{\boxed{\large\text{Answer: \huge \bf 7c + 2}}}\huge\checkmark\)

\(\text{Good luck on your assignment and enjoy your day!}\)

~\(\frak{Amphitrite1040:)}\)

Answer:

ok here is the answer

Step-by-step explanation:

how about you stop cheating and do it by your self

Answer:

V = 170 in³

Step-by-step explanation:

The volume of a triangular pyramid

V = (1/3)Ah

where A = area of the triangular base, and H = height of the pyramid

---------------------------

A = (1/2)(12)(5)

A = 30

V = (1/3)(30)(17)

V = 170 in³

The composite mapping (f o g)(x) ≥ 3 for the functions f(x) = 2x + 1 and g(x) = x² + 1. And (f o g)(x) > 0 for the functions f(x) = x² and g(x) = x.

A mapping is composite when the co- domain of the first mapping is the domain of the second mapping.

For the functions f(x) = 2x + 1 and g(x) = x² + 1

If x > 0, say x = 1 then;

g(x) = 1² + 1

g(x) = 2

(f o g)(x) = 2(2) + 1

(f o g)(x) = 4 + 1

(f o g)(x) = 5

If x = 0, then;

g(x) = 0² + 1

g(x) = 1

(f o g)(x) = 2(1) + 1

(f o g)(x) = 2 + 1

(f o g)(x) = 3

For the functions f(x) = x² and g(x) = x

If x < 0, say x = -0.5, then;

g(x) = -0.5

(f o g)(x) = (-0.5)²

(f o g)(x) = 0.25

In conclusion, the composite mapping (f o g)(x) ≥ 3 for functions f(x) = 2x + 1 and g(x) = x² + 1 and (f o g)(x) > 0 for the functions f(x) = x² and g(x) = x.

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Answer:

y = 3x + 4

Step-by-step explanation:

We can use the equation y = mx + b to find the equation:

Plug in the slope and the point into the equation, and we can find b:

-5 = 3(-3) + b

-5 = -9 + b

4 = b

Now, we can plug in the slope and y-intercept into the equation:

y = 3x + 4 will be our equation

Step-by-step explanation:

hope this help ! have a great day

The key features of the dashed graph are

A graph is a pictorial representation of a function

To describe the key features of the dashed graph, we proceed as follows.

So, the key features of the dashed graph are

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We first compute the partial derivatives of f(x, y):

f(x, y) = 5x² + 5y² - 10x + 21

∂f/∂x = 10x - 10

∂f/∂y = 10y

The critical points of f occur where both ∂f/∂x and ∂f/∂y are equal to zero. This only happens at the point (x, y) = (1, 0). (And this point does lie inside R.)

Next, compute the Hessian matrix H(x, y) for f :

\(H(x, y) = \begin{bmatrix}\frac{\partial^2f}{\partial x^2} & \frac{\partial^2f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2f}{\partial y^2}\end{bmatrix} = \begin{bmatrix}10&0\\0&10\end{bmatrix}\)

Since det(H(x,y)) = 100 is positive for all x and y, this means the critical point (1, 0) is a local minimum, and we have f(1, 0) = 16.

Next, we check for extrema along the boundary of R, which is comprised of a semicircle with radius 2 and the line segment connecting (-2, 0) and (2, 0).

• Parameterize the semicircular portion by

x(t) = 2 cos(t)

y(t) = 2 sin(t)

with 0 ≤ t ≤ π. Then

f(x(t), y(t)) = 20 sin²(t) + 20 cos²(t) - 20 cos(t) + 21

which is simplifies to a function of one variable t,

g(t) = 41 - 20 cos(t)

Find the extrema of g over the interval [0, π] : we have critical points when

g'(t) = 20 sin(t) = 0

which happens at t = 0 and t = π. At these points, we get local extreme values  of

•• t = 0   =>   x = 2 and y = 0   =>   f(2, 0) = 21

•• t = π   =>   x = -2 and y = 0   =>   f(-2, 0) = 61

• Over the line-segment portion, we take y = 0, so f(x, y) again reduces to a function of one variable:

f(x, 0) = 5x² - 10x + 21

Completing the square, we have

5x² - 10x + 21 = 5 (x - 1)² + 16

which has a maximum value of 16 when x = 1 (and this happens to be the critical point (1, 0) we found earlier).

So, over the region R, f(x, y) has

• an absolute maximum of 61 at the point (-2, 0), and

• an absolute minimum of 16 at (1, 0)

The total sum of the first 15 terms of the given arithmetic progression for a given problem is 1200.

initially, we have to discover the common difference

let the common difference for this question be d,

the primary term can be considered as a.

From the given information,

the whole of the 7th and 11th terms of an Arithmetic Progression is 35

At that point, the 7th term may be a + 6d, and the 11th term could be a + 10d.

and the summation of both terms is 35,

which gives the condition (a + 6d) + (a + 10d) = 35

reducing the condition, we get condition (i)

2a + 16d = 35 .......(i)

Essentially, the 9th term could ( a + 8d), and the 13th term is (a + 12d). and as per the given information, the entirety of the 9th and 13th terms is 65:    

which infers

(a + 8d) + (a + 12d) = 65

Streamlining this condition, we get:

2a + 20d = 65 ....(ii)

Presently, subtracting Condition (i) from Condition (ii), we get:

2a + 16d - 2a - 20d = 35 -65

-4d = -30

d = 7.5

Substituting this esteem of d into Condition (i) and Condition (ii), ready to fathom for a:

2a + 16(7.5) = 35

2a + 120 = 35

2a = -85

a = -42.5

In this manner, the common difference within the arithmetic progression is 7.5 and the first term is -42.5.

To discover the sum of the primary 15 terms of the arithmetic progression, we can use the equation:

S = (n/2)(2a + (n-1)d)

where S is the sum of the first n terms

a is the first term = -42.5.

d may be a common difference = 7.5

and n is the number of terms= 15

Substituting the values we have found, we get    S = (15/2)(2(-42.5) + (15-1)(7.5))

S = 15(-25 + 14(7.5))

S = 15(80)

S = 1200

Subsequently, the sum of the first 15 terms of the arithmetic progression is 1200.

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Answer:

I'll consider just 3

Step-by-step explanation:

To check if a number is a prime number it has to undergo series of division.

Let's not forget that a prime number is a number divisible by only itself and one

So for my argument.

2 cannot the number

3 cannot also divide the number

5 and 7 too cannot Divide the number.

I'll consider just 3 as the highest prime number to divide it

Answer:

yes

Step-by-step explanation:

simply substitute the x and y value coordinates in the equation.

4x-2y=6

4(1)-2(-1)=6

4+2=6

6=6

since it is true that 6=6,yes it is a solution

Step-by-step explanation:

The required shape is a square.

Refer to the attachment.

The sides of a square are equal. So, the answer is a square.

[Drawing Courtesy- Brainly app]

Answer:

$3.24

Step-by-step explanation:

Cost for the coffee + Cost for the bagel

$1.39 + $1.85

$3.24

Note that the coordinates of the point A' after rotating 90 degrees clockwise about the point (0,1) are (3, -4). (Option B)

To rotate a point 90 degrees  clockwise about a given point,we can follow these steps -

Translate the coordinates   of the given point so that the center of rotation is at the origin.   In this case,we subtract the coordinates   of the center (0,1) from the coordinates of point A (5,4) to get (-5, 3).

Perform the rotation by swapping the x and y coordinates and changing the sign of the   new x coordinate. In this case,we  swap the x and y coordinates of (-5, 3)   to  get (3, -5).

Translate the coordinates back to their original position   by adding the coordinates of the center (0,1)  to the result from step 2. In   this case, we add (0,1) to (3, -5) to get (3, -4).

Therefore, the coordinates   of the point A' after rotating 90 degrees clockwise about the point (0,1) are (3, -4).

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Answer:

Step-by-step explanation:

Image of H is H' with coordinates of:

On solving the provided question, we can say that -  here in graph we have on solving equation  \(2x^2+ 9y^2 \\\) = \(8 + 9 = 17\)

An equation is a formula in mathematics that joins two statements with the equal symbol = to represent equality. The definition of an equation in algebra is a mathematical statement proving the equality of two mathematical expressions. In the equation 3x + 5 = 14, for instance, the terms 3x + 5 and 14 are separated by an equal sign. The link between two phrases on either side of a letter is expressed mathematically. There is often only one variable, which is also the symbol. instance: 2x - 4 Equals 2.

here,

from graph, x= 2

y = 1

\(2x^2+ 9y^2 \\\)

\(8 + 9 = 17\)

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The money collected during the second week was :  840

Analysis:

Total number of persons.

number of persons that paid first week = 12

number that paid second week = 62 - 12 = 50

if each person is to pay 16.40, then 50 people would pay = 16.4 x 50 = 840 dollars.

Hence we can conclude that The money collected during the second week was :  840

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Answer:

  2.5%

Step-by-step explanation:

You want the percentage below 335 if the distribution is normal with a mean of 555 and a standard deviation of 110, using the empirical rule.

The z-score of 335 is ...

  Z = (X -µ)/σ

  Z = (335 -555)/110 = -220/110 = -2

The empirical rule tells you that 95% of the distribution is between Z = -2 and Z = 2. That is, 5% of the distribution is evenly split between the tails Z < -2 and Z > 2. Half that value is in each tail.

  P(X < 335) = 5%/2 = 2.5%

The percentage of people taking the test who score below 335 is 2.5%.

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The development of the personal budget that convinces parents that one can maintain an independent living is detailed as follows using the 50: 30: 20 rule:

50% for necessities (housing, food, transportation, utilities) $740

30% for luxuries, savings, vacations, entertainment, etc. $461

20% for Emergency Funds and Retirement $335

A personal budget is the household budget for a single person for a period.

The personal budget shows an estimate of the person's revenue and expenses over the period.

Hourly rate of earnings = $12

Working hours per week = 40 hours

Working hours per month = 160 hours (40 x 4 weeks)

Total monthly earnings = $1,920 ($12 x 160)

Assumed tax rate = 20%

After-tax take-home pay = $1,536 ($1,920 x 1 - 20%)

Rent of apartment = $300

Heating = $40

Water cost = $20

Sewage bill = $10

Food = $300

Transportation = $50

Other costs = $20

Total cost for necessities = $740

Thus, the development of the personal budget shows that one can maintain an independent living from the parents as a single person.

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The true statement is Statement A which is dogs of breed B typically weigh more than dogs of breed A using histogram in probability concept.

The likelihood of each possible occurrence is displayed on the y-axis of a probability histogram. A histogram is a statistical data display that makes use of rectangles to illustrate the frequency of data items in successive numerical intervals of equal size. The dependent variable is plotted along the vertical axis in the most typical type of histogram, while the independent variable is plotted along the horizontal axis. The distribution of data is depicted graphically in a histogram. A collection of neighboring rectangles, each with a bar that represents a certain type of data, is used to illustrate a histogram. Fill L1 with the values of the random variable (remember to use the class midpoints for continuous random variables). The associated percentages should be entered into L2 with a decimal format.

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