Some people say Mauna Kea is a taller mountain than Mt. Everest, because its total under sea and above sea level height is

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

7 0

Answer:

Mauna Kea has from the base that is to say the deepest 10 km of height that is to say 10,000 m which is equivalent to being higher than Mount Everest.jj

Step-by-step explanation:

The main reason why we could say that Mauna Kea is the highest mountain, even surpassing Mount Everest, is because of its altitude if we compare them we see the difference; for example Mauna Kea from its base that is to say from the deepest the ocean measures 10,000 m of altitude, on the other hand the Mount Everest has an altitude of 8848 meters (29,029 ft) above sea level, that is to say from its base it measures 8848 m altitude.

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Parker drove a total distance of 4 and 1 over 2 miles during the months of June and July. If Parker only drove 1 over 8 of a mil

never [62]

Answer: 4 and 1 over 2 ÷ 1 over 8

Step-by-step explanation:

Given that : Total distance drove by Parker =4 and 1 over 2 miles= $4\frac{1}{2}$ miles

Distance drove by Parker each day = 1 over 8 mile= $\frac{1}{8}$ mile

To find the total number of days , we need to divide the total distance by the distance drove by him in each day.

Thus, the number of days he went driving=4 and 1 over 2 ÷ 1 over 8

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3 years ago

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In a survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for th

marysya [2.9K]

Answer:

The population proportion is estimated to be with 99% confidence within the interval (0.1238, 0.2012).

Step-by-step explanation:

The formula for estimating the population proportion by a confidence interval is given by:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}$

Where:

$\hat{p}$ is the sample's proportion of success, which in this case is the people that regularly lie during surveys,

$z_{\alpha /2}$ is the critical value needed to find the tails of distribution related to the confidence level,

$n$ is the sample's size.

First we compute the $\hat{p}$ value:

$\hat{p}=\frac{successes}{n}=\frac{98}{603}=0.1625$

Next we find the z-score at any z-distribution table or app (in this case i've used StatKey):

$z_{\alpha /2}=2.576$

Now we can replace in the formula with the obtained values to compute the confidence interval:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}=0.1625\pm 2.576\times\sqrt{\frac{0.1625\times(1-0.1625)}{603}}=(0.1238, 0.2012)$