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amm1812
3 years ago
11

18. In their last basketball game, Holly scored

Mathematics
1 answer:
wariber [46]3 years ago
5 0

Answer:

Juanita's score was better

Step-by-step explanation:

Juanita scored 6.4 points above her average score whereas Holly only scored 2.5 points above her average score. (this was calculated via z-scores)

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Find the percent of change from the first value to the second. 20 ; 80
Lemur [1.5K]

Answer:

Percent Change Formula: [(new - old)/old] * 100

Step-by-step explanation:

New - old

80 - 20 = 60

Difference between new - old divided by old

60/20 = 3

Previous quotient times 100

3*100 = 300

Percent Change is 300%

Check your answer

300% of 20 is 60

20 + 60 = 80

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3 years ago
The equation r(t)= (2t)i + (2t-16t^2)j is the position of a particle in space at time t. Find the angle between the velocity and
ankoles [38]

Answer:

The answer is 135 degrees.

Step-by-step explanation:

As we are given the position. If we take the <u>derivative</u>, we get the velocity vector. If we take the <u>derivative</u> again, we find the acceleration vector of the particle.

r(t)=(2t)i+(2t-16t^{2})j

V(t)=2i+(2-32t)j

a(t)=-32j

At time t=0;

v(t)=2i+2j

a(t)=-32j

As i attach in the picture the angle between the velocity and acceleration vector is (45+90)=135 degrees

4 0
3 years ago
Estimate the following quantities without using a calculator. Then find a more precise result, using a calculator if necessary.
zubka84 [21]

Explanation:

a. It is so easy to double a number that no approximation is necessary.

  31,000 × 200 = 3.1×10⁴ × 2×10² = 6.2×10⁶ exactly

__

b. 6.2×10³ × 5.2×10⁶ ≈ 6×5×10⁹ = 3×10¹⁰ approximately

The approximation can be refined a bit by taking the ".2" into account:

  6.2×10³ × 5.2×10⁶ ≈ (6×5 + .2×(6+5))×10⁹

  = 32.2×10⁹ = 3.22×10¹⁰ approximately

Actual product: 3.224×10¹⁰.

For most purposes, the approximation is an adequate approximation, as it it within 10% of the actual value.

__

c. 9×10⁶ -2.3×10⁴ ≈ 9×10⁶ approximately

A better approximation is to actually subtract an approximation of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.02)×10⁶ ≈ 8.98×10⁶ approximately

The actual value uses all of the digits of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.023)×10⁶ = 8.977×10⁶ exactly

As in part B, either approximation is adequate for most purposes, as the difference from the actual value is less than .3%.

_____

The accuracy required of an approximation, hence the work you expend improving accuracy, should depend on the need in the final application of the number. Often, approximations are used for budget or resource planning purposes where some "slop" is allowed or even expected.

They can also be used in engineering applications, where the error needs to be on the side of more safety (rather than less).

7 0
3 years ago
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