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Papessa [141]
3 years ago
10

Please help me on this and thank you!

Mathematics
1 answer:
Lynna [10]3 years ago
7 0

Answer:

D

Step-by-step explanation:

We open it up, so 2/3x + -3 + -2 -1/3x, -3 + (-2) = -5.

2/3x - 1/3x - 5, 2/3x - 1/3x = 1/3x

Our final answer is 1/3x - 5

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2-(4-6x)+8=4x-(-2x-1)+5
klasskru [66]

Answer:

-4

Step-by-step explanation:

-4 but not really sure

5 0
3 years ago
Plz help, I need help!!!!
Otrada [13]

Answer:

56 multiplied by 9 = 504

Step-by-step explanation:

55.8 rounded is 56, and 9.12 rounded is 9

4 0
3 years ago
Battery packs in electric go-carts need to last a fairly long time. The run-times (time until it needs to be recharged) of the b
pav-90 [236]

Answer:

The minimum level for which the battery pack will be classified as highly sought-after class is 2.42 hours

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 2, \sigma = 0.33

What is the minimum level for which the battery pack will be classified as highly sought-after class

At least the 100-10 = 90th percentile, which is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{X - 2}{0.33}

X - 2 = 0.33*1.28

X = 2.42

The minimum level for which the battery pack will be classified as highly sought-after class is 2.42 hours

4 0
3 years ago
Anybody who knows Pre Calc?? My last offer here<br> Solution PLS!
4vir4ik [10]

Answers:

  • true speed = 26.935546 knots
  • true bearing = 102.265139 degrees

The values are approximate. Round them however you need to.

====================================================

Explanation:

The diagram you have drawn is correct if you're only considering one current. Specifically current #1 which has the bearing of 290 degrees moving at 25 knots.

The red angle is 290 degrees. The remaining portion that is needed to get a full 360 degrees is 360-290 = 70. This adds to the blue portion in quadrant 1, which is 90, and we get 70+90 = 160. So the blue angle is 160 degrees.

The vector for current #1 is <25*cos(160), 25*sin(160)>

This approximates to <-23.4923155, 8.5505036>

All of this is for current #1.

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

Vector #2 is much more simple in terms that there aren't messy decimal approximations here. The direction is due north, which means that the vector is pointing straight up toward 90 degrees.

We can then say: <12*cos(90), 12*sin(90)> = <12*0, 12*1> = <0, 12>

Vector #2 is <0,12>

When you wrote <12,0>, you were on the right track, but you had the coordinates swapped in the wrong order.

----------

Then we have to account for a wind going 4 knots in the southwest direction. This is at the angle 225 degrees (we rotate 180 degrees plus another 45 to get 225). So,

vector #3 = <4*cos(225), 4*sin(225)> = <-2.8284271, -2.8284271>

This vector's coordinates are approximate

---------

To recap everything so far, vectors 1 through 3 are the following

  • p = <-23.4923155, 8.5505036>
  • q = <0, 12>
  • r = <-2.8284271, -2.8284271>

I'm using variables to represent the vectors at this point, so we can then add them up to get...

s = p+q+r

s = <-23.4923155, 8.5505036> + <0,12> + <-2.8284271, -2.8284271>

s = <-23.4923155+0+(-2.8284271), 8.5505036+0+(-2.8284271)>

s = <-26.3207426, 5.7220765>

Vector s represents the sum of the three vectors p, q, and r. The two currents and the wind ultimately combine together to push the boat in the direction and along the length of vector s.

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

Now that we know the coordinates of vector s, we use them to find the true speed and true bearing of the boat.

speed = sqrt(a^2 + b^2)

speed = sqrt( (-26.3207426)^2 + (5.7220765)^2 )

speed = 26.935546 knots, which is approximate

angle = arctan(b/a)

angle = arctan(5.7220765/(-26.3207426))

angle = -12.265139

Add 360 to this angle to find that we get to the coterminal angle of -12.265139+360 = 347.734861 degrees

This angle is in quadrant 4, which is the southeast quadrant. If we rotate another 360-347.734861 = 12.265139 degrees, going counterclockwise, then we'll face directly east. Then another 90 degree counterclockwise rotation has us face directly north.

In total, the true bearing is approximately 90+12.265139 = 102.265139 degrees

So we'll start aiming north, and then rotate roughly 102.265139 degrees toward the east to get to the true bearing.

4 0
3 years ago
An aircraft carrier left Port 35 traveling north five hours before a container ship. The container ship traveled in the opposite
Molodets [167]

Answer:

Aircraft carrier speed = 4.97 km/h

Step-by-step explanation:

Given that the time taken by the aircraft = 5 hours

Let the distance covered = X

Speed = V

Time taken by the ship = 7 hours

Let the distance covered = Y

Speed = 8V

Since it is 8 km/h faster than the aircraft carrier.

Using the speed formula

Speed = distance/time

Distance = speed × time

For the aircraft;

X = 5V

For the ship:

Y = 7 × 8V = 56V

Since they are 303km distance apart,

X + Y = 303

5V + 56V = 303

61V = 303

V = 303/61

V = 4.97 km/h

Aircraft carrier speed = 4.97 km/h

5 0
3 years ago
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