2/3+1/2
HURRY I NED IT ANSWERED ASAP

Imagine that a survey of randomly selected people finds that people who used sunscreen were more likely to have been sunburned in the past year.

There are several possible explanations for the finding that people who used sunscreen were more likely to have been sunburned in the past year.

However, without additional context or data, it is difficult to determine the most likely explanation definitively. Here are a few possible explanations to consider:

1. Ineffective or improper use of sunscreen: It is possible that the individuals who reported using sunscreen did not apply it correctly or did not reapply it as recommended.

Inadequate application or failure to follow proper sunscreen usage guidelines could result in insufficient protection from the sun's harmful rays, leading to sunburn.

2. Self-selection bias: It is possible that individuals who have previously experienced sunburns may be more likely to use sunscreen. They may be more aware of the potential risks and take precautions, including using sunscreen.

This could create an association between sunscreen use and sunburn, even though sunscreen itself is intended to prevent sunburn.

3. Recall bias: The survey responses may be subject to recall bias, where individuals may inaccurately remember or report their sunscreen use and sunburn experiences.

Memory limitations or subjective perceptions of sunburn severity could influence the reported data and the observed association between sunscreen use and sunburn.

4. Confounding factors: There may be other factors or variables at play that are related to both sunscreen use and sunburn. For example, individuals who engage in activities that increase sun exposure or who have certain skin types may be more likely to both use sunscreen and experience sunburn.

It is important to note that further investigation, including more detailed surveys, data collection and statistical analysis, would be necessary to determine the most likely explanation for the observed association between sunscreen use and sunburn.

To know more about surveys refer here:

https://brainly.com/question/31685434#

#SPJ11

Answer:

66

Step-by-step explanation:

since BD and BC are congruent, angle C and angle D are equal. this means 6x-9=3x+24. If you solve this equation you get x=11. If you plug it into either 6x-9 or 3x+24 you get 57. 180-(57+57)=66

first one, the rest are nonsense

Answer:

Step-by-step explanation:  Hello!

I don't remember all of the postulates to these, but I hope this will help you! Vertical angles are congruent, so both angles SRT and PRQ are congruent.  This also means that line segments PQ and ST are congruent.  You also know that angles Q and S are congruent which is given.  By the ASA Theorem, when two angles and a side are congruent, then the triangles are congruent.

Answer:

\(17 \le t \le 28\)

Step-by-step explanation:

Given

\(t = 22\frac{1}{2}\) --- average time

\(\triangle t = 5\frac{1}{2}\) --- the variation

Required

The inequality to represent the scenario

To do this, we simply add and subtract the variation from the average time.

i.e.

\(t \± \triangle t\)

So, the inequality is:

\(22\frac{1}{2} - 5\frac{1}{2} \le t \le 22\frac{1}{2} + 5\frac{1}{2}\)

Solve:

\(17 \le t \le 28\)

Answer:

V = 133 cm³

Step-by-step explanation:

The formula for the volume of a cone is V = (πr²h) / 3, which is just the formula for the volume of a cylinder but divided by 3, by the way.

If the diameter is 6, divide that by 2 to get a radius of 3 cm. The height and pi are already given.

V = ((3.14)(3)²(12)) / 3

V = (339.12) / 3

V = 113.04 cm³, or if rounded, it should just be 133 cm³

Answer:

113.0

Step-by-step explanation:

You have to round. Also its correct on Imagine Math.

C. INFINITE SOLUTION CC

Answer:

no solution

Step-by-step explanation:

Answer:

greater than

Step-by-step explanation:

positive is higher than negative

Answer:

A,0

Beacuse 8×0=0

Answer:

3.) x= 1 y= -3    4.) x= 10 y= -10

Step-by-step explanation:

Answer:

no linear relationship exists between the two variables being compared

Step-by-step explanation:

The lengths of LV and OE are 15cm, and the lengths of LD and EV are 3√7 cm and 9/√7 cm, respectively.

Since LOVE is a kite, LV and OE are perpendicular bisectors of each other. Let the length of LD be x, and the length of EV be y. Then, we can use the Pythagorean theorem and the fact that the diagonals bisect each other to set up two equations:

x² + (LV/2)² = DV²/4

y² + (OE/2)² = LE²/4

Simplifying each equation and substituting the given values, we get:

x² + (LV/2)² = 81/4

y² + (OE/2)² = 225/4

We also know that the diagonals bisect each other, so we can set up another equation:

LV/2 + OE/2 = LO = VE

Substituting the given value for LE, we get:

LV/2 + OE/2 = 15

Solving this equation for one of the variables, we get:

LV = 30 - OE

Substituting this expression into the first equation above, we get:

x² + ((30 - OE)/2)² = 81/4

Simplifying and rearranging, we get:

OE² - 60OE + 675 = 0

Using the quadratic formula, we get:

OE = (60 ± √(3600 - 2700)) / 2

OE = 15 or 45

If OE = 15, then LV = 30 - 15 = 15, and we can solve for x and y:

x² + 7.5² = 81/4

y² + 7.5² = 225/4

Solving these equations, we get:

x = 3√7

y = 9/√7

If OE = 45, then LV = 30 - 45 = -15, which is impossible for a length. Therefore, the solution is:

LV = OE = 15

x = 3√7

y = 9/√7

Learn more about Pythagorean theorem here: brainly.com/question/14930619

#SPJ4

Complete question:

LOVE is a kite. LV and OE are diagonals. The segments DV = 9cm and LE = 15cm are given. Find the lengths of the other segments of the diagonals, DV, OE, and LV.

Answer:

\(y(t)=c_1e^{-t}+c_2te^{-t}+\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}\)

Step-by-step explanation:

Given the second-order differential equation. Solve by using variation of parameters.

\(y''+2y'+y=e^{-t}\ln(t)\)

(1) - Solve the DE as if it were homogeneous to find the homogeneous solution

\(y''+2y'+y=e^{-t}\ln(t) \Longrightarrow y''+2y'+y=0\\\\\text{The characteristic equation} \rightarrow m^2+2m+1=0, \ \text{solve for m}\\\\m^2+2m+1=0\\\\\Longrightarrow (m+1)(m+1)=0\\\\\therefore \boxed{m=-1,-1}\)

\(\boxed{\left\begin{array}{ccc}\text{\underline{Solutions to Higher-order DE's:}}\\\\\text{Real,distinct roots} \rightarrow y=c_1e^{m_1t}+c_2e^{m_2t}+...+c_ne^{m_nt}\\\\ \text{Duplicate roots} \rightarrow y=c_1e^{mt}+c_2te^{mt}+...+c_nt^ne^{mt}\\\\ \text{Complex roots} \rightarrow y=c_1e^{\alpha t}\cos(\beta t)+c_2e^{\alpha t}\sin(\beta t)+... \ ;m=\alpha \pm \beta i\end{array}\right}\)

Notice we have repeated/duplicate roots, form the homogeneous solution.

\(\boxed{\boxed{y_h=c_1e^{-t}+c_2te^{-t}}}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now using the method of variation of parameters, please follow along very carefully.

\(\boxed{\left\begin{array}{ccc}\text{\underline{Variation of Parameters Method(1 of 2):}}\\ \text{Given a DE in the form} \rightarrow ay''+by"+cy=g(t) \\ \text{1. Obtain the homogenous solution.} \\ \Rightarrow y_h=c_1y_1+c_2y_2+...+c_ny_n \\ \\ \text{2. Find the Wronskain Determinant.} \\ |W|=$\left|\begin{array}{cccc}y_1 & y_2 & \dots & y_n \\y_1' & y_2' & \dots & y_n' \\\vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \dots & y_n^{(n-1)}\end{array}\right|$ \\ \\ \end{array}\right}\)

\(\boxed{\left\begin{array}{ccc}\text{\underline{Variation of Parameters Method(2 of 2):}}\\ \text{3. Find} \ W_1, \ W_2, \dots, \ W_n.\\ \\ \text{4. Find} \ u_1, \ u_2, \dots, \ u_n. \\ \Rightarrow u_n= \int\frac{W_n}{|W|} \\ \\ \text{5. Form the particular solution.} \\ \Rightarrow y_p=u_1y_1+u_2y_2+ \dots+ u_ny_n \\ \\ \text{6. Form the general solution.}\\ y_{gen.}=y_h+y_p\end{array}\right}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(2) - Finding the Wronksian determinant

\(|W|= \left|\begin{array}{ccc}e^{-t}&te^{-t}\\-e^{-t}&e^{-t}-te^{-t}\end{array}\right|\\\\\Longrightarrow (e^{-t})(e^{-t}-te^{-t})-(te^{-t})(-e^{-t})\\\\\Longrightarrow (e^{-2t}-te^{-2t})-(-te^{-2t})\\\\\therefore \boxed{|W|=e^{-2t}}\)

(3) - Finding W_1 and W_2

\(W_1=\left|\begin{array}{ccc}0&y_2\\g(t)&y_2'\end{array}\right| \ \text{Recall:} \ g(t)=e^{-t} \ln(t)\\\\\Longrightarrow \left|\begin{array}{ccc}0&te^{-t}\\e^{-t} \ln(t)&e^{-t}-te^{-t}\end{array}\right|\\\\\Longrightarrow 0-(te^{-t})(e^{-t} \ln(t))\\\\\therefore \boxed{W_1=-t\ln(t)e^{-2t}}\)

\(W_2=\left|\begin{array}{ccc}y_1&0\\y_1'&g(t)\end{array}\right| \ \text{Recall:} \ g(t)=e^{-t} \ln(t)\\\\\Longrightarrow \left|\begin{array}{ccc}e^{-t}&0\\-e^(-t)&e^{-t} \ln(t)\end{array}\right|\\\\\Longrightarrow (e^{-t})(e^{-t} \ln(t))-0\\\\\therefore \boxed{W_2=\ln(t)e^{-2t}}\)

(4) - Finding u_1 and u_2

\(u_1=\int \frac{W_1}{|W|}; \text{Recall:} \ W_1=-t\ln(t)e^{-2t} \ \text{and} \ |W|=e^{-2t} \\\\\Longrightarrow \int\frac{-t\ln(t)e^{-2t}}{e^{-2t}} dt\\\\\Longrightarrow -\int t\ln(t)dt \ \text{(Apply integration by parts)}\\\\\\\boxed{\left\begin{array}{ccc}\text{\underline{Integration by Parts:}}\\\\uv-\int vdu\end{array}\right }\\\\\text{Let} \ u=\ln(t) \rightarrow du=\frac{1}{t}dt \\\\\text{an let} \ dv=tdt \rightarrow v=\frac{1}{2}t^2 \\\\\)

\(\Longrightarrow -\Big[(\ln(t))(\frac{1}{2}t^2)-\int [(\frac{1}{2}t^2)(\frac{1}{t}dt)]\Big]\\\\\Longrightarrow -\Big[\frac{1}{2}t^2\ln(t)-\frac{1}{2}\int (t)dt\Big]\\\\\Longrightarrow -\Big[\frac{1}{2}t^2\ln(t)-\frac{1}{2}\cdot\frac{1}{2}t^2 \Big]\\\\\therefore \boxed{u_1=\frac{1}{4}t^2-\frac{1}{2}t^2\ln(t)}\)

\(u_2=\int \frac{W_2}{|W|}; \text{Recall:} \ W_2=\ln(t)e^{-2t} \ \text{and} \ |W|=e^{-2t} \\\\\Longrightarrow \int\frac{\ln(t)e^{-2t}}{e^{-2t}} dt\\\\\Longrightarrow \int \ln(t)dt \ \text{(Once again, apply integration by parts)}\\\\\text{Let} \ u=\ln(t) \rightarrow du=\frac{1}{t}dt \\\\\text{an let} \ dv=1dt \rightarrow v=t \\\\\Longrightarrow (\ln(t))(t)-\int[(t)(\frac{1}{t}dt )] \\\\\Longrightarrow t\ln(t)-\int 1dt\\\\\therefore \boxed{u_2=t \ln(t)-t}\)

(5) - Form the particular solution

\(y_p=u_1y_1+u_2y_2\\\\\Longrightarrow (\frac{1}{4}t^2-\frac{1}{2}t^2\ln(t))(e^{-t})+(t \ln(t)-t)(te^{-t})\\\\\Longrightarrow\frac{1}{4}t^2e^{-t}-\frac{1}{2}t^2\ln(t)e^{-t}+ t^2\ln(t)e^{-t}-t^2e^{-t}\\\\\therefore \boxed{ y_p=\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}}\)

(6) - Form the solution

\(y_{gen.}=y_h+y_p\\\\\therefore\boxed{\boxed{y(t)=c_1e^{-t}+c_2te^{-t}+\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}}}\)

Thus, the given DE is solved.

Answer:

6(x-5)2+288

Step-by-step explanation:

6x2-60x+438  Factor out a 6 from the first 2 terms.

6(x2-10x     )+438  Do (b/2)2 to find the c term.  (-10/2)2 = 25

6(x2-10x+25)+438-150   Adding 25 inside the parenthesis is really like adding 6(25)=150 to that side of the equation.  To keep the equation in balance, you have to subtract 150 from the other term.  Simplify

6(x-5)2+288

Answer:

6x2-60x+438  Factor out a 6 from the first 2 terms.

6(x2-10x     )+438  Do (b/2)2 to find the c term.  (-10/2)2 = 25

6(x2-10x+25)+438-150   Adding 25 inside the parenthesis is really like adding 6(25)=150 to that side of the equation.  To keep the equation in balance, you have to subtract 150 from the other term.  Simplify

6(x-5)2+288

Step-by-step explanation:

Answer The formula df = (c – 1) (r – 1) is used to find the degree of freedom – The degree of freedom of the chi-square test is calculated by the formula df = (c – 1) (r – 1), where c represents the column number of cell and r represents the row number of the cell.

Hi there!  

»»————-　★　————-««

I believe your answer is:  

d – 22 ≥ 28

d ≥ 50  

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

\(\boxed{\text{Information Given:}}\\\\d - \text{Initial amount of money}\\\\\text{22 dollars was spent, and there are 'at least' 28 dollars left.}\)

⸻⸻⸻⸻\(\boxed{\text{Setting up an inequality:}}\\\\\text{22 dollars was spent from the initial amount, 'd'.}\\\\d - 22\\\\\text{There are \textbf{at least} 28 dollars left. "At least" indicates a \underline{greater than or equal to} sign.}\\\\\rightarrow \boxed{d-22\geq 28}\)

⸻⸻⸻⸻

\(\boxed{\text{Solving the inequality:}}\\\\d - 22\geq 28\\-------------\\\rightarrow d - 22 + 22 \geq 28 + 22\\\\\rightarrow \boxed{d \geq 50}\)

⸻⸻⸻⸻

\(\text{Your answer should be: }\boxed{d-22\geq 28; \text{ }d \geq 50}\)

⸻⸻⸻⸻

»»————-　★　————-««

Hope this helps you. I apologize if it’s incorrect.  

Answer:

terrells' age is 18 years old

Step-by-step explanation:

d represents dantes' age

t represents terrells' age

d= 3t

d+ t = 72

substitute d to the second equation

d+ t = 72

3t+t=72

4t = 72

t = 18

See my other articles for quiz and assignment in my site.

Just copy and paste "learningandassignments diy4pro" to google search engine.

Hope it helps.

The equation of the translated function is:

g(x) = (x + 3)^2 - 2

We can see that the blue graph is a translation of the red one, where the equation for the red graph is:

f(x) = x^2

Remember that the translations are:

Horizontal translation:

For a general function f(x), a horizontal translation of N units is written as:

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

If N is positive, the shift is to the left.

If N is negative, the shift is to the right.

Vertical translation:

For a general function f(x), a vertical translation of N units is written as:

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

If N is positive, the shift is upwards.

If N is negative, the shift is downwards.

By analyzing the vertices, we can see that the vertex of the blue graph is at (-3, -2) so we have a translation of 3 units to the left and 2 units down, meaning that we have:

g(x) = f(x + 3) - 2 = (x + 3)^2 - 2

So the correct option is A.

If you want to learn more about translations, you can read:

brainly.com/question/11468584

#SPJ1

A random sample taken with replacement from the original sample, and having the same size as the original sample, is known as a "bootstrap sample" or "bootstrap replication."

Bootstrapping is a resampling technique used to estimate the sampling distribution of a statistic. When we have a limited sample size and want to draw inferences about the population, we can use bootstrapping to create multiple resamples by randomly selecting observations from the original sample with replacement.

Here's how it works:

The key idea behind bootstrapping is that the original sample serves as a proxy for the population, and by repeatedly resampling from it, we can approximate the sampling distribution of the statistic of interest. This approach is especially useful when the underlying population distribution is unknown or non-normal.

By using a bootstrap sample that is the same size as the original sample, we maintain the same sample size and capture the variability present in the original data.

Learn more about bootstrapping here:

https://brainly.com/question/30778477

#SPJ11

\(\textbf{(a) Sample Mean, Sample Variance, and Sample Standard Deviation:}\)

To calculate the sample mean, sample variance, and sample standard deviation, we'll use the following formulas:

Sample Mean:

\(\[\bar{x} = \frac{{\sum_{i=1}^{n} x_i}}{n}\]\)

Sample Variance:

\(\[s^2 = \frac{{\sum_{i=1}^{n} (x_i - \bar{x})^2}}{n-1}\]\)

Sample Standard Deviation:

\(\[s = \sqrt{s^2}\]\)

where:

\(\(\bar{x}\) = sample mean,\\\(x_i\) = individual temperature measurements,\\\(n\) = number of measurements,\\\(s^2\) = sample variance,\\\(s\) = sample standard deviation.\)

Given the temperature measurements: 953, 950, 948, 955, 951, 955, we can plug these values into the formulas to calculate the desired statistics.

Sample Mean:

\(\[\bar{x} = \frac{953 + 950 + 948 + 955 + 951 + 955}{6} = 952\]\)

Sample Variance:

\(\[s^2 = \frac{(953-952)^2 + (950-952)^2 + (948-952)^2 + (955-952)^2 + (951-952)^2 + (955-952)^2}{6-1} = 6\]\)

Sample Standard Deviation:

\(\[s = \sqrt{6} \\\\\approx 2.449\)

Therefore, the sample mean is 952, the sample variance is 6, and the sample standard deviation is approximately 2.449.

\(\textbf{(b) Range and Median:}\)

The range is the difference between the largest and smallest values in the data set. The median is the middle value when the data set is arranged in ascending order.

Range:

The largest temperature measurement is 955, and the smallest is 948. Therefore, the range is \(\(955 - 948 = 7\)\).

Median:

Arranging the data set in ascending order: 948, 950, 951, 953, 955, 955. The middle value is the average of the two middle numbers, which are 951 and 953. Thus, the median is \(\(\frac{(951 + 953)}{2}= 952\)\).

The largest temperature measurement could increase without changing the median value by any amount less than or equal to half the range (\(\frac{7}{2}\) = 3.5 degrees Fahrenheit). As long as the largest temperature remains below 955 + 3.5, the median will remain unchanged.

To know more about Average visit-

brainly.com/question/18029149

#SPJ11

Answer:

b

Step-by-step explanation:

heavier ball by thw eight

The derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position is 2ax, as given by option b.

To find the position of maximum height of the pumpkin, the students need to find the point where the derivative of the vertical position with respect to the horizontal position is equal to zero. Setting 2ax equal to zero and solving for x, we get x=0. This means that the pumpkin reaches its maximum height at x=0, or in other words, at the point where it is launched from the trebuchet.

To find the value of the maximum height, we can substitute x=0 into the original equation for the pumpkin's trajectory. This gives us y(0) = b, which means that the maximum height of the pumpkin is b units.

The derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position is 2ax because the derivative of ax^2 with respect to x is 2ax. This means that the rate of change of the pumpkin's height with respect to its horizontal position is proportional to 2ax. When x is zero, the derivative is also zero, which indicates that the pumpkin has reached its maximum height at that point.

This is because at the maximum height, the rate of change of height with respect to horizontal distance is zero. Finally, we find the value of the maximum height by substituting x=0 into the equation for the pumpkin's trajectory.

Learn more about derivative:

brainly.com/question/29144258

#SPJ11

The distance, which is determined by integrating the velocity function over the specified time interval, is calculated as 2.98 and 2.58.

We will use left Riemann sum and right Riemann sum to obtain an overestimate and an underestimate. The definite integral, which in this case is a distance, can also be approximated using the average of these estimates.

1 ) The distance the bug crawls obtained by integrating the velocity function v(t) = 4/1+t over the inclined time interval is

\(\int\limits^1_0 \, \frac{4}{1+t} dt.\)

Splitting the interval [0, 1] into sub intervals of length Δt = 0.2, we get the Riemann sum L and the right Riemann sum R:

L = ∑⁴ v(ti) Δt

L = (0.2) (v(0) + v(0.2) + v(0.4) + v(0.6) + v(0.8))

L ≈ 2.98

R = ∑⁵ v(ti) Δt

R = (0.2) (v(0.2) + v(0.4) + v(0.6) + v(0.8) + v(1.0))

R ≈ 2.58

2 )The average of these two Riemann sums gives us a new estimate for the distance :

AVG = (2.98 + 2.58)/2

=2.78.

Learn more about Riemann sums :

https://brainly.com/question/29275224

#SPJ4

Answer:

The function is decreasing for x > -1

Step-by-step explanation:

Answer:

9 worms are longer than 3 inches

Step-by-step explanation:

you can just count how many are in in the rows after 3 and that is your answer

9. The distribution used to determine if the coefficients of a regression are significantly different from zero is the t-distribution. The t-distribution is commonly used in hypothesis testing for regression analysis, where the null hypothesis is that the coefficient is equal to zero.

10. Without a specific scatterplot provided, it is not possible to determine the relationship between x and y. To provide an answer, I would need more information or a description of the scatterplot.

Visit here to learn more about coefficients brainly.com/question/1594145

#SPJ11

Answer: 29, 31, 36, 38...

Step-by-step explanation: The pattern is +5 and then +2

Answer: 29

Step-by-step explanation:

The pattern in this sequence is:

5 + 2 + 5 and so on

So you add 24 + 5 which equals 29

I hope I helped you with this question. Have a great day!

Step-by-step explanation:

hlejsjsisjdjsisjsjsjsjsjsisjsnsjssjjsjsjs as