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3 years ago

A sailboat traveled 30 miles in 2.5 hours. At the same speed, how far can the boat sail in another 4.5 hours ?

Mathematics

1 answer:

Ivanshal [37]3 years ago

8 0

54 miles

using the formula

speed = $\frac{distance}{time}$ = $\frac{30}{2.5}$ = 12 mph

continuing at 12 mph then

distance = speed × time = 12 × 4.5 = 54 miles

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Ian wants to run 412 miles to get ready for his upcoming race. He has set up 3 markers the same distance apart to show how far h

Salsk061 [2.6K]

Answer:

Each marker will be exactly <u>103</u> miles apart.

Step-by-step explanation:

It is given that Ian wants to run 412 miles and has set up 3 markers the same distance apart.

Between the start and end of run, the 3 separate markers will divide the 412-mile distance into 4 equal segments.

So each segment = 412/4 = 103 miles

Each marker will be exactly <u>103</u> miles apart.

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3 years ago

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Gracelyn has multiple sticks of different lengths to build a triangle for a class project. Which 3 lengths could Gracelyn choose

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Hello

11 cm, 25 cm, 15 cm
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Step-by-step explanation:

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3 years ago

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2 years ago

Medical scientists study the effect of acute infection on tissue-specific immunity. In a collection of experiments under the sam

Serjik [45]

Answer:

The 95% confidence interval for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

In a collection of experiments under the same conditions, 44 of 75 mice test positive for lymphadenopathy. This means that $n = 75$ and $\pi = \frac{44}{75} = 0.586$ .

Compute a 95% confidence interval for the true proportion of mice that will test positive under similar conditions.

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 - 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.5291$

The upper limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 + 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.6429$

The 95% confidence interval for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).

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3 years ago