All categories

3 years ago

Please hurry...

Mathematics

2 answers:

Natasha_Volkova [10]3 years ago

6 0

Answer:

Which formula can be used to find the surface area of the hemisphere? ... Recall that the formula for the volume for a sphere is v=4/3πr^3 and the ... Type A is a spherical ball with a radius of 12 inches. ... r=6cm. SA=1/2(4πr^2)+πr^2. SA=2π(6)^2+π(6)^2. SA=72(3.14)+36(3.14) ... The diameter of the hemisphere is 48 feet.

Step-by-step explanation:

DochEvi [55]3 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

Solve by Elimination: 8x + 6y = -5 10x + 6y= -13

jonny [76]

Answer:

x= -4 and y= 27/6

Step-by-step explanation:

-(8x + 6y = -5) which converts to -8x -6y = 5

10x + 6y = -13

simplify from there

-8x + 10x = 2x ; -6y + 6y = 0 ; 5 - 13 = -8

soo, now you have

2x = -8

x = -4

then, plug in to find y

8(-4) + 6y = -5

-32 + 6y = -5 add 32 on both sides

6y = 27 divide both sides by 6

y= 27/6 or 4.5

8 0

3 years ago

True or False: You can tell if two ratios form a proportion if their cross products are equal. True O False

daser333 [38]

Answer:

Step-by-step explanation:

False

7 0

2 years ago

Please keep as simple as possible

Alenkinab [10]

Answer:

f(x) = (X^3 +6) + 4

Step-by-step explanation:

to move left add +6 from inside the parenthesis

to move up add +4 outside of parenthesis

*since there were no parenthesis you add them*

7 0

3 years ago

Read 2 more answers

Could someone answer this pls thank u

loris [4]

Given:

The area model.

The area of the shape is 82 square units.

To find:

The correct equation for the given situation.

Solution:

We know that, area of a rectangle is

$Area=Length \times Width$

The figure is divided into 3 rectangles. Using this formula, the area of each rectangle is

$A_1=(x)\times (3x)=3x^2$

$A_2=(x)\times (4)=4x$

$A_3=(3)\times (3x)=9x$

Now, the total area is equation to 82 square units.

$A_1+A_2+A_3=82$

$3x^2+4x+9x=82$

$3x^2+13x=82$

Since the required equation is $3x^2+13x=82$ , therefore, the correct option is C.

7 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 $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

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

7 0

3 years ago