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Inessa05 [86]
3 years ago
10

Tom drives a truck. His regular trip is a distance of 280 km. He drives at an average speed of 80 km/h.

Mathematics
1 answer:
ivolga24 [154]3 years ago
3 0

Answer:

.5 hour, or 30 minutes

Step-by-step explanation:

280 km / 80 kmh = 3.5 hr before limiter

280 km / (80-10) kmh = 4 hr after

so .5 hour from 3.5 to 4 hours. or 30 minutes

This is provided he matains this speed at all times.

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Hello there, to answer your question...

258, percentage increased by 43% (percent) of its value

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I hope this information helps.

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5+(-5)=0

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Step-by-step explanation:

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In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
pickupchik [31]

Answer:

The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus 11 - 1 = 10.

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

  • The area to the left of the lower end of the interval shall be 1/2 \cdot (1 - 0.90)= 0.05.
  • The area to the left of the upper end of the interval shall be 0.05 + 0.90 = 0.95.

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

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  • a cumulative probability of 0.95.

t \approx 1.812.

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

\displaystyle t \cdot \frac{s_{n-1}}{\sqrt{n}},

where

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  • s_{n-1} is the unbiased estimate for the standard deviation, and
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For this confidence interval:

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  • s_{n-1} = 14, and
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Hence the width of the 90% confidence interval is

\displaystyle 1.812 \times \frac{14}{\sqrt{10}} \approx 7.65.

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

(38.3, 53.7).

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