Ask question

Ask question

All categories

3 months ago

14

Jorden drives to the store at 30 miles per hour. On her way home she averages only 20 miles per hour. If the total driving time

takes half an hour, how far does she live from the store?

2 answers:

babunello [11.8K]3 months ago

7 0

Response:

Rates are 30, 20.

Distance is x miles each way.

Detailed explanation:

Svet_ta [12.7K]3 months ago

4 0

For this scenario, we can define:
$d = v * t
$
Where,
d represents distance
v signifies speed
t indicates time
Rearranging to find time gives us:
$t = d / v
$
The total driving duration is half an hour:
$t = d / v1 + d / v2

1.5 = d / 30 + d / 20$
Rearranging for d results in:
$1.5 * 30 * 20 = 20d + 30d

1.5 * 30 * 20 = 50d

d = (1.5 * 30 * 20) / (50)

d = 18 miles$
Conclusion:
she lives 18 miles from the store

You might be interested in

WNAE, an all-news AM station, finds that the distribution of the lengths of time listeners are tuned to the station follows the

Inessa [12570]

Answer:

a) $P(X>20)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{20-15}{3.5})=P(z>1.43)$

The probability can be determined using the complement rule, with the standard normal distribution, an excel sheet, or a calculator.

$P(z>1.43)=1-P(z$

b) $P(X$

This probability can also be calculated using the normal standard distribution, an excel sheet, or a calculator.

$P(z$

c) $P($

For this one, the probability can likewise be derived from the standard normal distribution, excel, or a calculator, with specific adjustments:

$P(-1.43$

Step-by-step explanation:

Previous concepts

Normal distribution refers to a symmetric probability distribution centered around the mean, indicating that occurrences near the mean are more common than those far from it.

The Z-score is a statistic that represents a value's relationship to the average of a set of values, expressed in terms of how many standard deviations it is away from the mean.

Part a

Let X denote the random variable representing the lengths within a population, and for our case, the distribution for X is as follows:

$X \sim N(15,3.5)$

Where $\mu=15$ and $\sigma=3.5$

We seek the probability:

$P(X>20)$

The most effective way to solve this is by leveraging the normal distribution and the corresponding Z-score:

$z=\frac{x-\mu}{\sigma}$

By applying this formula, we can find the probability:

$P(X>20)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{20-15}{3.5})=P(z>1.43)$

Again, this probability can be obtained either using the complement rule, the standard normal distribution, or a calculator.

$P(z>1.43)=1-P(z$

Part b

$P(X$

This probability can also be computed using either the normal standard distribution, an excel sheet, or a calculator.

$P(z$

Part c

$P($

In this case, the probability can similarly be acquired with the help of the standard normal distribution, an excel sheet, or a calculator, with particular adjustments:

$P(-1.43$

5 0

3 months ago

The table shows data from Vicente's Form 1040. Gross income $75,000 Standard deduction $12,000 Taxable income $63,000 Total inco

Inessa [12570]

a+b+c = d indicates that a and d are equal.

7 1

3 months ago

Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

PIT_PIT [12445]

Answer:

Step-by-step explanation:

The equation representing the sphere, which has its center at the origin, can be written as $x^2+y^2+z^2 = 64$ . For z equal to 4, we find

$x^2+y^2= 64-16 = 48$ .

This results in a circle with a radius of $4\sqrt[]{3}$ in the x-y plane.

c) We will build on the analysis from earlier to set limits in both Cartesian and polar coordinates. Initially, we recognize that x spans from $-4\sqrt[]{3}$ to $4\sqrt[]{3}$ . This determination is made by fixing y = 0 and identifying the extreme x values that fall on the circle. For y, we observe that it ranges between $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , which holds because y must reside within the interior of the identified circle. Lastly, z will extend from 4 up to the sphere; hence, it varies from 4 to $\sqrt[]{64-x^2-y^2}$ .

The respective triple integral representing the volume of D in Cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Remember that the cylindrical coordinates are expressed as $x=r\cos \theta, y = r\sin \theta,z = z$ , where r denotes the radial distance from the origin projected onto the x-y plane. Also note that $x^2+y^2 = r^2$ . We will derive new limits for each of the transformed coordinates. Recall that due to the prior circular constraint, $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ varies between 0 and $2\pi$ . Furthermore, r starts from the origin and extends to the edge of the circle, with r reaching a maximum of 4\sqrt[]{3}. Lastly, Z increases from the plane z=4 up to the sphere, where it is constrained by \sqrt[]{64-r^2}. Thus, the integral that computes the desired volume is as follows:

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . It’s important to note that the r factor arises from the Jacobian associated with the transition from Cartesian to polar coordinates, ensuring the integral maintains its value. (Explaining how to calculate the Jacobian exceeds the scope of this response).

a) When dealing with spherical coordinates, keep in mind that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ , where $\phi$ denotes the angle formed between the vector and the z axis, varying from 0 to pi. It is crucial to recognize that at z=4, this angle remains constant along the circle we previously identified. Let’s determine the angle by selecting a point on the circle and employing the angle formula between two vectors. Setting z=4 and x=0 gives us y=4 $\sqrt[]{3}$ by taking the positive square root of 48. We will now compute the angle between the vector $a=(0,4\sqrt[]{3},4)$ and vector b =(0,0,1), which represents the unit vector along the z axis. We apply the following formula

$\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}$

Consequently, across the circle, $\phi = \frac{\pi}{3}$ . Observe that rho transitions from the plane z=4 to the sphere, with rho reaching up to 8. Given $z = \rho \cos \phi$ , we have that $\rho = \frac{4}{\cos \phi}$ at the plane. Thus, the corresponding integral is

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta$ , where the new factor incorporates the Jacobian for the spherical coordinate system.

d) Let’s work with the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$ .

It’s important to observe that the integral can be separated since the inner part remains independent of theta. By implementing the substitution $u = 64-r^2$ , we achieve $\frac{-du}{2} = r dr$ , leading to

$=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}$

3 0

2 months ago

1. The following are the number of hours that 10 police officers have spent being trained in how to handle encounters with peopl

zzz [12365]

Answer:

$Range = 16$

$Inter\ Quartile\ Range = 6.75$

$Variance = 20.44$

$Standard\ Deviation = 4.52$

Step-by-step explanation:

Provided

4, 17, 12, 9, 6, 10, 1, 5, 9, 3

Calculating the range;

$Range = Highest - Lowest$

From the data provided;

The highest value is 17 and the lowest is 1

Thus;

$Range = 17 - 1$

$Range = 16$

Calculating the inter-quartile range

The inter-quartile range (IQR) is computed as follows

$IQR = Q_3 - Q_1$

Where

Q3 = Upper Quartile and Q1 = Lower Quartile

Start by sorting the data in ascending order

1, 3, 4, 5, 6, 9, 9, 10, 12, 17

N = Total data points; N = 10

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

Calculating Q3

$Q_3 = \frac{3}{4}(N+1) th\ item$

Substituting 10 for N

$Q_3 = \frac{3}{4}(10+1) th\ item$

$Q_3 = \frac{3}{4}(11) th\ item$

Expressing 8.25 as 8 + 0.25 $Q_3 = \frac{33}{4} th\ item$

$Q_3 = 8.25 th\ item$

Converting 0.25 into fractions

$Q_3 = (8 + 0.25) th\ item$

$Q_3 = 8th\ item + 0.25 th\ item$

From the ordered dataset;

$Q_3 = 8th\ item +\frac{1}{4} th\ item$ and

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

$8th\ item = 10$ $9th\ item = 12$

$Q_3 = 8th\ item +\frac{1}{4} (9th\ item - 8th\ item)$

$Q_3 = 10 +\frac{1}{4} (12 - 10)$

$Q_3 = 10 +\frac{1}{4} (2)$

$Q_3 = 10 +0.5$

$Q_3 = 10.5$

Calculating Q1

Substituting 10 for N

$Q_1 = \frac{1}{4}(N+1) th\ item$

Expressing 2.75 as 2 + 0.75

$Q_1 = \frac{1}{4}(10+1) th\ item$

$Q_1 = \frac{1}{4}(11) th\ item$

$Q_1 = \frac{11}{4} th\ item$

$Q_1 = 2.75 th\ item$

Converting 0.75 into fractions

$Q_1 = (2 + 0.75) th\ item$

$Q_1 = 2nd\ item + 0.75 th\ item$

From the arranged dataset;

and $Q_1 = 2nd\ item +\frac{3}{4} th\ item$

$Q_1 = 2nd\ item +\frac{3}{4} (3rd\ item - 2nd\ item)$

$2nd\ item = 3$ $3rd\ item = 4$

$Q_1 = 3 +\frac{3}{4} (4 - 3)$

$Q_1 = 3 +\frac{3}{4} (1)$

$Q_1 = 3 +0.75$

$Q_1 = 3.75$

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

Remember that

$IQR = Q_3 - Q_1$

$IQR = 10.5 - 3.75$

$IQR = 6.75$

Calculating Variance

Begin by determining the mean

$Mean = \frac{1+3+4+5+6+9+9+10+12+17}{10}$

$Mean = \frac{76}{10}$

$Mean = 7.6$

Then subtract the mean from each value and square the differences

$(1 - 7.6)^2 = (-6.6)^2 = 43.56$

$(3 - 7.6)^2 = (-4.6)^2 = 21.16$

$(4 - 7.6)^2 = (-3.6)^2 = 12.96$

$(5 - 7.6)^2 = (-2.6)^2 = 6.76$

$(6 - 7.6)^2 = (-1.6)^2 = 2.56$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(9 - 7.6)^2 = (1.4)^2 = 1.96$

$(10 - 7.6)^2 = (2.4)^2 = 5.76$

$(12 - 7.6)^2 = (4.4)^2 = 19.36$

$(17 - 7.6)^2 = (9.4)^2 = 88.36$

Sum up the squared results

$43.56 + 21.16 + 12.96 + 6.76 + 2.56 + 1.96 + 1.96 + 5.76 + 19.36 + 88.36 = 204.4$

Then divide that sum by the total number of observations;

$Variance = \frac{204.4}{10}$

$Variance = 20.44$

Calculating Standard Deviation (SD)

$SD = \sqrt{Variance}$

$SD = \sqrt{20.44}$

$SD = 4.52$ (Approximated)

4 0

2 months ago