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3 years ago

What is the range for the amount of hours studied? 6 18 4 12

Mathematics

1 answer:

Ne4ueva [31]3 years ago

5 0

Answer:

Step-by-step explanation:

You take the most amount studied (18) and the least amount studied (12) and subtract them and that would be the range

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Suppose you choose a team of two people from a group of n > 1 people, and your opponent does the same (your choices are allow

jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

$\left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right i^{3}$

Step-by-step explanation:

selecting the first team from n people we have $\left(\begin{array}{ccc}n\\1\\\end{array}\right) = n$ possibility and choosing second team from the rest of n-1 people we have $\left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1$

As { A, B} = {B , A}

Therefore, the total possibility is $\frac{n(n-1)}{2}$

Since our choices are allowed to overlap, the second team is $\frac{n(n-1)}{2}$

Possibility of choosing both teams will be

$\frac{n(n-1)}{2} * \frac{n(n-1)}{2} \\\\= [\frac{n(n-1)}{2}] ^{2}$

We now have the formula

1³ + 2³ + ........... + n³ = $[\frac{n(n+1)}{2}] ^{2}$

1³ + 2³ + ............ + (n-1)³ = $[x^{2} \frac{n(n-1)}{2}] ^{2}$

= $\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] = [\frac{n(n-1)}{2}]^{3}$

4 0

4 years ago

The diameter of a sphere is 16 cm. What is the sphere's volume? Round to the nearest tenth.

ohaa [14]

Answer:

V ≈ 2144.70 $cm^3$

Step-by-step explanation:

Radius is going to be used instead of the diameter. Half of 16 is 8, and so that will be the value that shall be used.

$V = (4/3)\pi (8)^3$

V ≈ 2144.66 $cm^3$

V ≈ 2144.70 $cm^3$

5 0

3 years ago

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 i

stich3 [128]

Complete question :

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 inches. Round answers to the nearest tenth of a percent. a) What percentage of years will have an annual rainfall of less than 43 inches? b) What percentage of years will have an annual rainfall of more than 39 inches? c) What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

Answer:

0.61053

0.66314

0.26647

Step-by-step explanation:

Given that :

Mean (m) = 41.4 inches

Standard deviation (s) = 5.7 inches

a) What percentage of years will have an annual rainfall of less than 43 inches?

P(x < 43)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (43 - 41.4) / 5.7 = 0.2807017

p(Z < 0.2807) = 0.61053 ( Z probability calculator)

b) What percentage of years will have an annual rainfall of more than 39 inches? c)

P(x > 39)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (39 - 41.4) / 5.7 = −0.421052

p(Z > −0.421052) = 0.66314 ( Z probability calculator)

What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

P(x < 38)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (38 - 41.4) / 5.7 = −0.596491

p(Z < −0.596491) = 0.27545 ( Z probability calculator)

P(x < 42)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (42 - 41.4) / 5.7 = 0.1052631

p(Z < 0.1052631) = 0.54192 ( Z probability calculator)

0.54192 - 0.27545 = 0.26647

6 0

3 years ago

raheel travelled 3654 km by bus and train. if he went 2365 km by bus, how far did he travel by train?

gladu [14]

3654 - 2365 = 1289 km

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krok68 [10]

Can you make the picture a little clearer

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3 years ago

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