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KengaRu [80]
3 years ago
6

A group of band instructors answered a survey about hours of rehearsal per week and number of competitions won. The graph shows

the results of this survey.
Based on these results, if a band practices 5 hours per week next season, which is the best estimate of the number of competitions the band can expect to win?
A.
36
B.
31
C.
27
D.
24
Mathematics
2 answers:
Irina18 [472]3 years ago
8 0

Answer:

its B

Step-by-step explanation:

i took the quiz on PLATO

IRISSAK [1]3 years ago
5 0

Answer: B. 31

Step-by-step explanation:

This linear regression was constructed by relating the hours practiced per week and the number of competitions won.

Going by this graph, the number of competitions they can expect to win at 5 practices a week is 31.

This is derived by looking for the point where 5 competitions on the x-axis intersects with the line. This point is at 31 competitions on the y axis which would make it the answer.

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Answer:

Step-by-step explanation:

5. a) ∠1 and ∠2 are remote interior angles of ∠ACD so that means that ∠ACD = ∠1 + ∠2

   b) Because an exterior angle is the sum of its two remote interior angles it makes sense that an exterior angle is greater in measure than either of its remote interior angles.

6. BD = DB  Reflexive property

    ∠3 = ∠5, ∠4 = ∠6  Alt. int. angles

    ΔADB = ΔCDB   ASA

7. AB = BC Def. of midpoint

   ∠1 = ∠2 Given

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   ΔBAE = ΔCBD ASA

    ∠D = ∠E CPCTC

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3 years ago
Express 5/8 as a percent.<br> 1.40%<br> 2.62.5%<br> 3.58%<br> 4.62%
Drupady [299]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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The circumference of the equator of a sphere was measured to be 82 82 cm with a possible error of 0.5 0.5 cm. Use linear approxi
True [87]

Answer:

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.

Step-by-step explanation:

The circumference (s), in centimeters, and the surface area (A_{s}), in square centimeters, of a sphere are represented by following formulas:

A_{s} = 4\pi\cdot r^{2} (1)

s = 2\pi\cdot r (2)

Where r is the radius of the sphere, in centimeters.

By applying (2) in (1), we derive this expression:

A_{s} = 4\pi\cdot \left(\frac{s}{2\pi} \right)^{2}

A_{s} = \frac{s^{2}}{\pi^{2}} (3)

By definition of Total Differential, which is equivalent to definition of Linear Approximation in this case, we determine an expression for the maximum error in the calculated surface area (\Delta A_{s}), in square centimeters:

\Delta A_{s} = \frac{\partial A_{s}}{\partial s} \cdot \Delta s

\Delta A_{s} = \frac{2\cdot s\cdot \Delta s}{\pi^{2}} (4)

Where:

s - Measure circumference, in centimeters.

\Delta s - Possible error in circumference, in centimeters.

If we know that s = 82\,cm and \Delta s = 0.5\,cm, then the maximum error is:

\Delta A_{s} \approx 8.3083\,cm^{2}

The maximum error in the calculated surface area is approximately 8.3083 square centimeters.

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3 years ago
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