All categories

3 years ago

Which is the side length of a cube with a volume of 2744 cm3? Please help

Mathematics

2 answers:

Tasya [4]3 years ago

7 0

The volume of the cube is a^3, where a is the length
to find out the length you need to do
∛2744=14
a=14

Elis [28]3 years ago

3 0

$\sf~V=a^3$

Plug in what we know:

$\sf2744=a^3$

Find the cube root of both sides:

$\sf~a=\sqrt[3]{2744}$

Simplify:

$\sf~a=\boxed{\sf14}$

You might be interested in

Place the steps for finding f'(X) in the correct order.<br><br><br> Plz help <br> Thank you!

SashulF [63]

Answer:

See below

Step-by-step explanation:

Given function is ,

=> f(x) = 4x + 7.

Replace f(x) with y ,

=> y = 4x + 7

For finding the inverse , interchange x and y ,

=> x = 4y + 7

Now solve for y ,

=> 4y = x - 7

=> y = x-7/4

Replace y with f-¹(x) ,

=> f-¹(x) = (x-7)/4

These are the required steps .

8 0

2 years ago

A researcher wants to test whether a certain sound will make rats do worse on learning tasks. It is known that an ordinary rat c

AveGali [126]

Answer:

So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.

Step-by-step explanation:

In this data we have

Mean= u = 18

X= 38

Standard deviation = s= 6

1) We formulate the null and alternate hypothesis as

H0: u = 18 against Ha : u > 18 One tailed test .

2) The significance level alpha = ∝= 0.05 and Z alpha has a value ± 1.645 for one tailed test.

3)The test statistics used is

Z= X- u / s

z= 38-18/6= 3.333

4) The calculated value of z = 3.33 is greater than the z∝ = 1.645

5) So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.

First we set the criteria for determining the true of value of the variable. That whether the rats learn in less or more than 18 trials.

Then we find the value of z for the given significance value given and the test about to be checked.

Then the test statistic is determined and calculated.

Then both value of z and z alpha re compared. If the test statistics falls in the rejection region reject the null hypothesis and conclude alternate hypothesis is true.

The figure shows that the calulated z value lies outside the given z values

4 0

3 years ago

The length of a rectangle is 2 less than twice the width. The perimeter of the rectangle is 38. Find the length and the width of

hjlf

Answer:

YEET A DOG

Step-by-step explanation:

YEETING A DOG GIVES U COOL POINTS LOLOLOLOLOLOLOLOLOLOLOL

6 0

3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$