Ask question

Ask question

All categories

2 years ago

13

the volume is 50.24 cubic inches and has a radius of 4 inches. What is the height? Use 3.14 for pie.

1 answer:

Dmitrij [34]2 years ago

8 0

Volume = (1/3)(pi)(r^2)(h)
= (1/3)*3.14*(4^2)(h)
= 16.747h

volume / 16.747 = height = 50.24/16.747 = 3

You might be interested in

Decimal greater than 1 write the word form and expanded form for each 3.4

True [87]

3.4, 3 + 0.4, three point four
5.8, 5 + 0.8, five point eight
6.2, 6 + 0.2, six point two

5 0

2 years ago

A right triangle has legs measuring 6 and 7 inches. What is the length of the hypotenuse?

Alchen [17]

Answer:

5

Step-by-step explanation:

8 0

3 years ago

Seventy cards are numbered 1 through 70 , one number per card. One card is randomly selected from the deck. What is the probabil

Vanyuwa [196]

Answer:

$Probability = \frac{2}{35}$

Step-by-step explanation:

Given

$Total = 70$

First, we need to list the multiples of 5

$M_5 = \{5,10,15,20,25,30,35,40,45,50,55,60,65,70\}$

Then, multiples of 3 $M_3 = \{3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69\}$

Next, is to list out the common elements in both

$M_3\ n\ M_5 = \{15,30,45,60\}$

$n(M_3\ n\ M_5) = 4$

The required probability is then calculated as thus:

$Probability = \frac{n(M_3\ n\ M_5)}{Total}$

$Probability = \frac{4}{70}$

$Probability = \frac{2}{35}$

7 0

3 years ago

Arrange the entries of matrix A in increasing order of their cofactors values

givi [52]

To find the cofactor of

$A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]$

We cross out the Row and columns of the respective entries and find the determinant of the remaining $2\times 2$ matrix with the alternating signs.

$Ac_{11}=\left|\begin{array}{ccc}4&-1\\2&1\end{array}\right|$

$Ac_{11}=4\times 1- -1\times 2$

$Ac_{11}=4+ 2$

$Ac_{11}=6$

$Ac_{12}=-\left|\begin{array}{ccc}-7&-1\\-8&1\end{array}\right|$

$Ac_{12}=-(-7\times 1- -1\times -8)$

$Ac_{12}=-(-7- 8)$

$Ac_{12}=15$

$Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|$

$Ac_{21}=-(5\times 1- 3\times 2)$

$Ac_{21}=-(5-6)$

$Ac_{21}=1$

$A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|$

$Ac_{23}=-(7\times 2 -8\times 5)$

$Ac_{23}=-(14-40)$

$Ac_{23}=26$

$A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|$

$Ac_{31}=5\times -1 -4\times 3$

$Ac_{31}=-5-12$

$Ac_{31}=-17$

$A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|$

$Ac_{33}=7\times 4- -7\times 5$

$Ac_{33}=28+35$

$Ac_{33}=63$

Therefore in increasing order, we have;

$Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63$

7 0

3 years ago

Read 2 more answers

Out of 120 high school boys and 80 high school girls, 45 tried out for track. Of those 45, 12 were girls.

zhannawk [14.2K]

<span>So there are 33 boys who tried out for track. And this is 27.5 % of the total boys. And there 15% of the girls who tried out for the track. And in total 22.5 % of the total population tried out for track</span>

5 0

2 years ago