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xxMikexx [17]
2 years ago
15

Ruby has a DRF of 1.95 and a 6-month basic rate of $450. What is her annual

Mathematics
2 answers:
insens350 [35]2 years ago
6 0
The answer on this question is D
harkovskaia [24]2 years ago
3 0

Answer:

For me it was A sooooo

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A company claims that its tablet computers have an average recharge time of 3 hours. In a random sample of these computers, the
jek_recluse [69]

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μ < 3

Step-by-step explanation:

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1 year ago
A sequence is defined recursively using the formula f(n + 1) = –0.5 f(n) . If the first term of the sequence is 120, what is f(5
balu736 [363]
Hello,

f(1)=120\\&#10; f(2)=-\frac{1}{2}*120=-60\\ &#10; f(3)=-\frac{1}{2} *(-60)=30\\ &#10; f(4)=-\frac{1}{2} *(30)=-15\\ &#10; f(5)=-\frac{1}{2} *(-15)=\frac{15}{2}\\&#10;
3 0
2 years ago
Aidan rode his bike 2.3 km from home to the library.Then he biked to the park.When he left home, his odometer read 293.8 km.When
Vinvika [58]

Answer:

308.2 - 293.8 = 14.4 km

14.4 - 2.3 = 12.1

So the library is 12.1 km from the park.

4 0
2 years ago
13/14 as a decimal rounded to the nearest tenth
Ratling [72]

Answer: .93

Step-by-step explanation: .9, .93, or .929

3 0
2 years ago
Read 2 more answers
A survey of 225 students showed the mean number of hours spent studying per week was 20.6 and the standard deviations was 2.7
LekaFEV [45]

Answer:

The margin of error is approximately 0.3

Step-by-step explanation:

The following information has been provided;

The sample size, n =225 students

The sample mean number of hours spent studying per week = 20.6

The standard deviation = 2.7

The question requires us to determine the margin of error that would be associated with a 90% confidence level. In constructing confidence intervals of the population mean, the margin of error is defined as;

The product of the associated z-score and the standard error of the sample mean. The standard error of the sample mean is calculated as;

\frac{sigma}{\sqrt{n} }

where sigma is the standard deviation and n the sample size. The z-score associated with a 90% confidence level, from the given table, is 1.645.

The margin of error is thus;

1.645*\frac{2.7}{\sqrt{225}}=0.2961

Therefore, the margin of error is approximately 0.3

5 0
2 years ago
Read 2 more answers
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