Ask question

Ask question

All categories

3 years ago

10

Celia filled the glasses shown below completely with water. The total amount of water that Celia poured into the glasses is 80 c

ubic centimeters. What is the height of glass 1? Round your answer to the nearest tenth. (Use π = 3.14.) Note that all measurements are in centimeters and images are not drawn to scale.

2 answers:

bagirrra123 [75]3 years ago

7 0

The height of glass 1 is 7.6
so, you find the volume of glass 2,= 26.376
then you subtract the answer from 80 cubic cm to get the volume of glass 1
80 - 26.376 = 53.694
then you work backwards to get the height, which is 7.6

WITCHER [35]3 years ago

5 0

Answer:

The height of glass 1 is 7.6 cm.

Step-by-step explanation:

We can see that glass 1 is a cylinder whose volume is given by :

$V=\pi r^{2} h$

its r = $3/2=1.5$

h = ?

The second glass is a cone with h = 6.3 and r = $4/2=2$

Volume is given as :

$\pi r^{2} \frac{h}{3}$

=> $3.14\times2\times2\times\frac{6.3}{3}$

V = 26.38 cubic cm

Given is :

The total amount of water that Celia poured into the glasses is 80 cubic centimeters.

This means she poured $80-26.38=53.62$ cubic cm of water in cylindrical glass. Solving for 'h'

$53.62=3.14\times1.5\times1.5\times h$

$7.065h=53.62$

h = 7.6 cm

Therefore, the height of glass 1 is 7.6 cm.

You might be interested in

Find the perimeter of the

spayn [35]

D is the right answer

8 0

2 years ago

ASAP! GIVING BRAINLIEST! Please read the question THEN answer CORRECTLY! NO guessing. Show your work OR give an explaination.

Leya [2.2K]

Answer:

y=1/2x+9

Step-by-step explanation:

By the least steep graph, it means the "smallest" slope. The smaller the slope is, the less steep it'd be. The slope can be positive or negative, too.

1/4x<-1/2x<2x<-6x

Therefore, y=1/2x+9 would be the equation with the least steep graph.

I hope this helped! Have a good day!

8 0

3 years ago

- Which of the following expressions has a coefficient of 10 and a constant of 5?

maw [93]

Answer:

Either 10+5 or 10-5

Step-by-step explanation:

Coefficient- the number that is multiplied by the variable

Constant- a number that stays the same and is not changed no matter what x or y equals.

6 0

2 years ago

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

Which 30 60 90 triangle is labled with the correct side length ratio

Evgesh-ka [11]

Isosceles triangle.

6 0

2 years ago