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3 years ago

Please help! I’ll give brainlist...!

Mathematics

2 answers:

nika2105 [10]3 years ago

8 0

Answer: second option -300+150+(-50)

Explanation:

She going down, the value is negative. When going up, the value is positive. So when the swimmer swims down 300 feet, the value is negative. Then he goes up 150 ft, which makes the value positive. Finally, he swims down 50 ft, so the valued is negative.

Ronch [10]3 years ago

5 0

Answer:

B. -300 + 150 + (-50)

Step-by-step explanation: If a swimmer is diving down that takes him below ground level which is why he'd be going negative. therefore your answer would be -300 + 150 + (-50).

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Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The

lys-0071 [83]

Answer:

The 85% confidence interval for the population proportion that claim to always buckle up is (0.7877, 0.8441).

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 level of $1-\alpha$ , 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 $1 - \frac{\alpha}{2}$ .

They randomly survey 391 drivers and find that 319 claim to always buckle up.

This means that $n = 391, \pi = \frac{319}{391} = 0.8159$

85% confidence level

So $\alpha = 0.15$ , z is the value of Z that has a pvalue of $1 - \frac{0.15}{2} = 0.925$ , so $Z = 1.44$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8159 - 1.44\sqrt{\frac{0.8159*0.1841}{391}} = 0.7877$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8159 + 1.44\sqrt{\frac{0.8159*0.1841}{391}} = 0.8441$

The 85% confidence interval for the population proportion that claim to always buckle up is (0.7877, 0.8441).

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Answer:

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