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Romashka-Z-Leto [24]

3 years ago

15

Your uncle lives 20 miles away. So far, your family has driven 6 miles toward your uncle's house. How

2 answers:

patriot [66]3 years ago

6 0

14 miles 20-6!!! Good luck

Harrizon [31]3 years ago

5 0

Answer:

14 miles

Step-by-step explanation:

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A circular test track for cars has a circumference of 2.7km . A car travels around the track from the southernmost point to the

Galina-37 [17]

This problem is about the difference between distance and displacement.
The distance is how many kilometers has the car covered, the displacement is how far from the original point is the car now, and does not depend on the path followed, but it is the shortest path that connects two points.

a) In this part, they ask you for the distance. From the southernmost to the northernmost point of the circumference, the car has traveled half of the circumference, therefore:
distance = circumference ÷ 2 = 2.7km ÷ 2 = 1.35km

b) In this part, you are asked for the displacement, which in this case is the diameter of the circumference:
displacement = diameter = circumference ÷ π = 2.7km ÷ 3.14 = 0.86km = 860m

5 0

3 years ago

Convert the integral below to polar coordinates and evaluate the integral.

White raven [17]

I suppose the integral is

$\displaystyle\int_0^5\int_y^{\sqrt{25-y^2}} xy\,\mathrm dx\,\mathrm dy$

The integration region corresponds to a sector of a cirlce with radius 5 subtended by a central angle of π/4 rad. We can capture this region in polar coordinates by the set

$R=\left\{(r,\theta)\mid0\le r\le5,0\le\theta\le\dfrac\pi4\right\}$

Then $x=r\cos\theta$ , $y=r\sin\theta$ , and $\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$ . So the integral becomes

$\displaystyle\iint_Rxy\,\mathrm dx\,\mathrm dy=\int_0^{\pi/4}\int_0^5r^3\sin\theta\cos\theta\,\mathrm dr\,\mathrm d\theta=\frac{625}{16}$

7 0

3 years ago

When Noah goes bowling, his scores are normally distributed with a mean of 150 and a standard deviation of 12. What is the proba

Zarrin [17]

Answer:

The probability of his score being between 135 and 167 is 0.8151 or (0.8151*100=81.51%)

Step-by-step explanation:

Given that:

Mean = μ = 150

SD = σ = 12

Let x1 be the first data point and x2 the second data point

We have to find the z-scores for both data points

x1 = 135

x2 = 167

So,

$z_1 = \frac{x_1-mean}{SD}\\= \frac{135-150}{12}\\=\frac{-15}{12}\\=-1.25$

And

$z_2 = \frac{x_2-mean}{SD}\\z_2 =\frac{167-150}{12}\\=\frac{17}{12}\\= {1.416}$

We have to find area to the left of both points then their difference to find the probability.

So,

Area to the left of z1 = 0.1056

Area to the left of z2 = 0.9207

Probability to score between 135 and 167 = z2-z1 = 0.9027-0.1056 = 0.8151

Hence,

The probability of his score being between 135 and 167 is 0.8151 or (0.8151*100=81.51%)

3 0

3 years ago

Use this graph of f to find f(6).

Ludmilka [50]

Answer: I will help you once you add a graph in the comments of this post

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A test was given to a group of students. The grades and gender are summarized below A B C TotalMale 5 9 2 16Female 7 11 12 30Tot

faust18 [17]

Probability that the student got a 'C' GIVEN they are female = number of females that got a C in the test/number of females

From the information given,

number of females that got a C in the test = 12

number of females = 30

Thus,

Probability that the student got a 'C' GIVEN they are female = 12/30

We would simplify the fraction by dividing the numerator and denominator by 6. Thus,

Probability that the student got a 'C' GIVEN they are female = 2/5

4 0

1 year ago