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Lynna [10]
3 years ago
11

A pendulum is constructed by attaching a 4m rope to a bowling ball. It is then displaced 2m to one side and released.

Mathematics
1 answer:
Arada [10]3 years ago
8 0

Answer:

T = 4.0 s

Step-by-step explanation:

The period of a simple pendulum is given by

period = 2\pi \sqrt{ \frac{l}{g} }

where l = length of the pendulum

= 4 m

g = acceleration due to gravity

= 9.8 m/s^2

period = 2\pi \sqrt{ \frac{(4 \: m)}{(9.8 \:  \frac{m}{ {s}^{2} } )} }  = 4.0 \: s

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For this question the answer would be x=-3 and x=-2
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When josh planted his bean plant, it stood only 2 inches tall. Since then it has grown 3.25 inches every two weeks. Write the eq
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Answer:

y=3.25x+2

Step-by-step explanation:

Y is the total height after x weeks and since it has grown 3.25 inches per 2 weeks it would be 3.25x and add 2 because the initial height is 2 inches

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Find 10 partial sums of the series. (round your answers to five decimal places.) ∞ 15 (−4)n n = 1
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Given

\Sigma_{n=1}^\infty15(-4)^n

The first 10 partial sums are as follows:

S_1=\Sigma_{n=1}^{1}15(-4)^n=15(-4)=\bold{-60} \\  \\ S_2=\Sigma_{n=1}^{2}15(-4)^n=\Sigma_{n=1}^{1}15(-4)^n+15(-4)^2 \\ =-60+15(16)=-60+240=\bold{180} \\  \\ S_3=\Sigma_{n=1}^{3}15(-4)^n=\Sigma_{n=1}^{2}15(-4)^n+15(-4)^3 \\ =180+15(-64)=180-960=\bold{-780} \\  \\ S_4=\Sigma_{n=1}^{4}15(-4)^n=\Sigma_{n=1}^{3}15(-4)^n+15(-4)^4 \\ =-780+15(256)=-780+3,840=\bold{3,060} \\  \\ S_5=\Sigma_{n=1}^{5}15(-4)^n=\Sigma_{n=1}^{4}15(-4)^n+15(-4)^5 \\ =3,060+15(-1,024)=3,060-15,360=\bold{-12,300}

S_6=\Sigma_{n=1}^{6}15(-4)^n=\Sigma_{n=1}^{5}15(-4)^n+15(-4)^6 \\ =-12,300+15(4,096)=-12,300+61,440=\bold{49,140}

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7 0
3 years ago
Of all sixth graders, 70% sent a text message yesterday. Ten trials of a simulation are conducted and the data are recorded belo
Effectus [21]

Answer:

80%

Step-by-step explanation:

According to the given data, the numbers 0 through 6 represent students who sent a text yesterday while the numbers 7 through 9 represent students who did not send a text yesterday.

Based on this key, we will calculate the number of students who sent the text yesterday and the number of students who did not sent the text yesterday.

The given data is: 62072, 34570, 80983, 04292, 83150, 36330, 96268, 14077, 77985, 13511

For each group we will identify how many students sent the text and how many students did not. According to the key, a number from 0 to 6 will be counted as a student who sent the text and a number from 7 to 9 will be counted as a student who did not send the text. Students who did not sent the text are made bold in the groups below.

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34570:  4 students sent the text, 1 did not

80983:  2 students sent the text, 3 did not

04292:  4 students sent the text, 1 did not

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We need to find the probability that 3 or more of a group of 5 students randomly selected will send a text today.

From the given groups, the number of groups where 3 or more students sent a text = 8

This represents our number of favorable or desired outcomes.

Total number of possible outcomes is the total number of groups = 10

Since, probability is defined as the ratio of number of favorable outcomes to total number of outcomes, based on the simulated data  the probability that 3 or more of a group of 5 students randomly selected will send a text today will be = \frac{8}{10}=0.80=80\%

Thus, based on the simulated data there is 80% probability that 3 or more of a group of 5 students randomly selected will send a text today

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3 years ago
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