Juan and Gloria are comparing the two numbers 354.6 and 354.06. Juan says that 354.06 is equal to 10 x 354.6. Gloria says he is

"A sample of 20 randomly chosen water melons was taken from a large population, and their weights were measured. The mean weight

AnnZ [12381]

Answer: (97.98, 112.020)

Step-by-step explanation: We will create a 95% confidence interval for the average weight of melons.

Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

Sample mean = x = 105 lb

Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

tα/2 represents the critical value for the test (which will be determined using the t distribution table).

To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

</pfor>

6 0

2 months ago

How many times greater is the value of the 3 in 4,367 than the value of the 3 in 39?

Inessa [12570]

Approximately 111.97

8 0

1 month ago

A flat circular plate has the shape of the region x2 + y2≤1. The plate, including the boundary where x2 + y2 = 1, is heated such

Leona [12618]

Setting both partial derivatives to zero results in a single critical point at $(x,y)=\left(\dfrac12,0\right)$ , located within the unit disk.

At this given point, the derivative value of the Hessian matrix is

$|H|=\begin{vmatrix}T_{xx}&T_{xy}\\T_{yx}&T_{yy}\end{vmatrix}=\begin{vmatrix}2&0\\0&4\end{vmatrix}=8>0$

and the second-order partial derivative with respect to $x$ yields

$T_{xx}\bigg|_{(x,y)=(1/2,0)}=2>0$

This suggests that the critical point represents a local minimum, marking it as the coldest area on the plate with a temperature of $T\left(\dfrac12,0\right)=-\dfrac14$ .

To find the hottest area on the plate, it must be located along the boundary. Let $x=\cos\theta$ and $y=\sin\theta$ , so that

$T(x,y)=T(\theta)=\cos^2\theta+2\sin^2\theta-\cos\theta$
$T(\theta)=\dfrac32-\cos\theta-\dfrac12\cos2\theta$

Thus, the plate's boundary (the circle $x^2+y^2=1$ ) is treated as a single variable function $\theta$ examined over $\theta\in[0,2\pi)$ . A single differentiation gives

$T'(\theta)=\sin\theta+\sin2\theta=0$
$\implies\theta=0,\theta=\dfrac{2\pi}3,\theta=\pi,\theta=\dfrac{4\pi}3$

You will discover that $T(\theta)$ achieves three extrema on the interval $(0,2\pi)$ , with relative maxima occurring at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$ , and a relative minimum at $\theta=\pi$ (and $\theta=0$ , if you wish to include that).

Our minimum has already been identified inside the plate - which you can check to have a lower temperature than at the points noted by $T(\theta)$ - and we identify two maxima at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$ , both showing a maximum temperature of $T=\dfrac94$ .

Reverting to Cartesian coordinates, these points match up with $\left(-\dfrac12,\pm\dfrac{\sqrt3}2\right)$ .

4 0

1 month ago

A student solved the problem below by first dividing 20 by 10. What mistake did the student make? A baseball team has won 20 gam

Leona [12618]

Answer:

Step-by-step explanation: The error made by the student was dividing the wins by the losses.

The student should have divided the wins by the total number of games played.

Initially, the student ought to have summed 20 and 10 to conclude there were 30 games in total.

8 0

2 months ago

Read 2 more answers

Xander says that it is not possible to draw a quadrilateral that is not a parallelogram and not a trapezoid. Devin says it is po

Zina [12379]

To address this question effectively, it's essential to clarify the characteristics of the given geometric terms.

A quadrilateral is defined as a four-sided figure.

Parallelograms are a subset of quadrilaterals characterized by having two pairs of parallel sides, with congruent opposite sides and angles. Instances of parallelograms include squares, rectangles, and rhombuses.

Trapezoids are also categorized under quadrilaterals but are distinguished by having only one pair of parallel sides.

In our investigation, we seek to determine whether there exists a four-sided shape without parallel sides. The included image presents an example of a quadrilateral that does not qualify as either a parallelogram or a trapezoid. Therefore, in accordance with the definitions of these polygons, the correct choice is B.

6 0

1 month ago