All categories

3 years ago

I am thinking of a number. It is less than 500. It’s cube root is an integer. What is the largest possible value of this number

Mathematics

1 answer:

myrzilka [38]3 years ago

4 0

Answer:

$343$

Step-by-step explanation:

$1^{3} =1\\2^{3} =8\\3^{3} =27\\4^{3} =64\\5^{3} =125\\6^{3} =216\\7^{3} =343\\8^{3} =512$

$\sqrt[3]{x } >500$

$\sqrt[3]{343} >500$

You might be interested in

Use set notation to describe the venn diagram. Here's the picture of the diagrams!! https://imgur.com/a/G5oCszI

Nikolay [14]

The first picture represents the elements in B that are not (also) in A. So, we want the set difference of B and A:

$B \setminus A = \{x \in B:\ x \notin A\}$

For the second picture, we can work like this: if we take all the elements belonging to A, we only need to add the bottom-right part, which is the intersection of B and C. So, we can express the blue part as

$A \cup (B\cap C) = \{x:\ x \in A \lor (x\in B \land x\in C) \}$

7 0

4 years ago

Jenna has bought a new hybrid car. Each week for a period of seven weeks, she has noted the fuel efficiency (in miles per gallon

Leviafan [203]

Answer: 0.82 miles per gallon

Step-by-step explanation:

Given : Jenna has bought a new hybrid car. Each week for a period of seven weeks, she has noted the fuel efficiency (x) (in miles per gallon) of her car.

The results are :45 44 43 44 45 44 43

The formula to find the standard deviation is given by :-

$\sigma=\sqrt{\dfrac{\sum_{i=1}^n(x_i-\overline{x})^2}{n-1}}$

First we find mean of the given data ,

$\overline{x}=\dfrac{\sum^n_{i=1}x_i}{n}\\\\=\dfrac{45+44+43+44+45+44+43}{7}=\dfrac{308}{7}=44$

Now,

${\sum_{i=1}^n(x_i-\overline{x})^2=(1)^2+(0)^2+(-1)^2+(0)^2+(1)^2+(0)^2+(-1)^2$

$=1+1+1+1=4$

Now, standard deviation of the above results will be :-

$\sigma=\sqrt{\dfrac{4}{7-1}}\\\\=\sqrt{\dfrac{4}{6}}\\\\=\dfrac{2}{\sqrt{3}}=0.816496580928\approx0.82$

Hence, the standard deviation of these results = 0.82 miles per gallon

8 0

4 years ago

What are the zeros of the function below? Check all that apply.

geniusboy [140]

Pretty sure it’s D and F because if you set (x-4)=0, x is 4 and if you set (x+7)=0, x is -7

3 0

3 years ago

Read 2 more answers

Find the product of (2/3)x and (1/2x²)+6x-3

kvasek [131]

Answer:

x2+

x−3

Step-by-step explanation:

3 0

3 years ago

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In e

Sedaia [141]

The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let $\%R_{A}$ be the quantity of nonzero elements in $R_{A}.$

Let $C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right)$ .

Parity demands that $\%R_{A}$ and $\%R_{B}$ must equal 2 or 4.

Case 1: $\%R_{A}$ =4 and $\%R_B$ =4. There are $\left(\begin{array}{l}3\\ 2\end{array}\right)$ =3 ways to put 2-1's in $R_A$ , so there are 3 ways.

Case 2: $\%R_{A}$ =2 and $\%R_B$ =4. There are 3 ways to position the -1 in $R_A$ , 2 ways to put the remaining -1 in $R_B$ (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of $\%R_{A}=4,\%R_{B}=2$ for a complete of 24 ways.

Case 3: $\%R_A=\%R_B=2$ . There are 3 ways to put the -1 in $R_{A}$ . Now, there are two cases on what happens next.

The 1 in $R_B$ goes directly under the -1 in $R_A$ . There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in $R_C$ and $R_D$ . (Either the 1 comes first in $R_C$ or the 1 comes first in $R_D$ .)
The 1 in $R_B$ doesn't go directly under the -1 in $R_A$ . There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

#SPJ4

3 0

2 years ago

Read 2 more answers