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3 years ago

A pelican starts at 60 feet above sea level. It descends 60 feet to catch a fish. How do I explain this?

2 answers:

8 0

Answer: 60 + (-60) = 0
explanation: The Pelican who is 60 feet above sea level is represented by +60.Now, he is trying to catch the fish which is at 0 level, means Sea level because the pelican is trying to catch the fish by descending 60 feet

scoray [572]3 years ago

6 0

Answer:

The answer is 0

Step-by-step explanation:

Because its original height was 60 feet above sea level but it descends to 60 feet to capture fish so 60 - 60 = 0

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Step-by-step explanation:

75-30=45

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3 years ago

4 + {−5 + [−3 + 4 + 2(−7 + 4) + 4] + 2}

scoray [572]

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Step-by-step explanation:

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2 years ago

Find the angle of depression from the top of a lighthouse, 260 feet above water to a ship that's 270 feet offshore?

expeople1 [14]

Answer: OPTION C.

Step-by-step explanation:

Observe the triangle ABC attached.

Notice that the angle of depression is represented with $\alpha$ .

Knowing that the top of a lighthouse is 260 feet above water and the ship is 270 feet offshore, you can find the value of $\alpha$ by using arctangent:

$\alpha= arctan(\frac{opposite}{adjacent})$

In this case you can identify that:

$opposite=260\\adjacent=270$

Therefore, substuting values into $\alpha= arctan(\frac{opposite}{adjacent})$ , you get that the angle of depression is:

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3 years ago

A teacher was interested in knowing the amount of physical activity that his students were engaged in daily. He randomly sampled

klasskru [66]

Answer:

The standard error of the mean is 4.5.

Step-by-step explanation:

As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:

$\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}$

The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.

The mean of the sample is:

$\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5$

The standard deviation of the sample is calculated as:

$s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8$

Then, we can calculate the standard error of the mean as:

$\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}=\frac{26.8}{6}= 4.5$

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3 years ago