Which size random sample is likely to provide the most trustworthy results

A football player gained two yards on one play on the next play he gained 5 ft was his gain greater on the first play or the sec

saveliy_v [14]

The gain was greater on the first play.

1 yard=3 ft
2yards times 3= 6ft

6ft>5ft

Hope this is helpful!

4 0

3 years ago

Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)

garik1379 [7]

The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

$\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})$

Find its derivative:

$\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})$

Now its magnitude:

$\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}$

After simplifying:

$\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}$

Now the unit tangent is:

$\rm T(u) = \dfrac{R'(t)}{|R'(t)|}$

After dividing and simplifying, we get:

$\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)$

Now, finding the derivative of T(u), we get:

$\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})$

Now finding its magnitude:

$\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)$

After simplifying, we get:

$\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}$

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

$\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})$

Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

Learn more about the vector here:

brainly.com/question/8607618

#SPJ4

3 0

2 years ago

Can someone please help with these problems I’m having trouble pls

ololo11 [35]

When the chords intersect inside the circle the sum of the vertical angles and the sum of the referenced arcs are the same. Since the vertical angles are congruent, each is half the sum of the referenced arcs.

Similarly when the chords intersect outside the circle, the angle is half the <em>difference</em> of the arc measures.

It's the same with a tangent; in that case there's just not a part of the arc that's not counted on the side with the tangent.

OK, that's how. Let's try some.

1. x = (88 + 86)/2 = 87

2. x = (150 - 60)/2 = 45

3. The angle is exactly half the subtended arc,

x = 110/2 = 55

4. The referenced arc is 360 - 220 = 140 degrees and x is half, 70.

5. Outside, half the difference, x = (90-20)/2 = 35

6. Angle outside, inner arc 6 degrees, vertical angle is 90 degrees.

90 = (x - 6)/2

180 = x - 6

x = 186

7. There's a special rule for this case but we can treat it as outside, little arc x, big arc 360-x,

60 = (360 - x - x)/2

120 = 360 - 2x

2x = 240

x = 120

8. Inner arc x, outer arc is 360-140-x = 220-x. Half the difference is 38,

38 = (220 - x - x)/2

76 = 220 - 2x

2x = 220 - 76

x = (220 - 76)/2 = 72

9. x is formed by a tangent and a radius, so a right angle, x=90

y is subtended by a 35 degree angle so y=70

10. Remaining triangle angle is 180 - 62 - 28 = 90. Arc that subtends it is double, 180. That means the long side is a diameter.

11. Remaining triangle angle is 180 - 50 - 70 = 60. x=120, y=100, z=140

12. Remaining arc 360 - 170 - 90 = 100. Angles are half.

x=45, y=50, z=85

3 0

3 years ago

7. You purchased crackers for $2.79, a steak for $5.27. ice cream for

bixtya [17]

Answer:

no who pays almost 9 bucks for ice cream

3 0

4 years ago

Read 2 more answers

Can anyone help me find the area ty

vredina [299]

The area is 39.69 I hope this helped!

4 0

2 years ago