Please help me with number 2 (15pts)

What is the volume of a cone with a radius of 4 centimeters and a height of 9 centimeters? Use 3.14 to estimate p.

lilavasa [31]

The volume is 150.8 cm³

6 0

3 years ago

Read 2 more answers

Can someone please please please help with this problem preferably by the end of today? (later is fine too tho)

V125BC [204]

Answer:

Add all of the items, then multiply by 5

Step-by-step explanation:

Why by 5? Well, in order to make

5 cookies per player one batch would be made

so, 10 cookies pp = 2 batches

15 cookies pp = 3 batches

20 cookies pp = 4 batches

and then you would need 4 more to make 2 dozen, so you would then make another batch and have a few extra, but you would have 2 dozen per player!

hope this helps!!

3 0

2 years ago

Help pls

babymother [125]

Answer:

1). x = 10 m

2). x = 15 cm

3). x = 5 yd

4). AB = 10 units

Step-by-step explanation:

1). By Pythagoras theorem in the given triangle,

a² + b² = c²

Where 'c' = Hypotenuse

a and b = Legs of the right triangle

By substituting measures of the sides in the formula,

x² = 8² + 6²

x = $\sqrt{100}$

x = 10 m

2). By using Pythagoras theorem in this triangle,

x² = 9² + (12)²

x² = 81 + 144

x = $\sqrt{225}$

x = 15 cm

3). By Pythagoras theorem,

(13)² = x² + (12)²

169 = x² + 144

169 - 144 = x²

25 = x²

x = 5 yd

4). If BD is a perpendicular bisector of AC,

AD = CD = 6 cm

By Pythagoras theorem in ΔABD,

AB² = BD² + AD²

AB² = 8² + 6²

AB = $\sqrt{100}$

AB = 10 units

4 0

3 years ago

Read 2 more answers

Factor completely x2 − 36

Hoochie [10]

X² - 36 = x² - 6² = (x-6)(x+6)

5 0

2 years ago

One thousand independent rolls of a fair die will be made. Compute an approximation to the probability that the number 6 will ap

pashok25 [27]

Answer:

0.19

Step-by-step explanation:

We are given that

Number of rolls of die=n=1000

Let the event of six coming up be success.Then, in each trial , the probability of success =p=P(success)=P(6)= $\frac{1}{6}$

Let X be the random variable for the number of sixes in the 1000 rolls of die.

Then, $X\sim Binom(1000,\frac{1}{6})$

Since, n is very large,the binomial random variable can be approximated as normal random variable.

Mean, $\mu=np=1000\times \frac{1}{6}=166.67$

Variance= $\sigma^2=np(1-p)=1000\times \farc{1}{6}\times (1-\frac{1}{6})=1000\times \frac{5}{36}=138.89$

$X\sim N(166.67,138.89)$

$P(150\leq X\leq 200)=P[\frac{150-166.67}{11.79}\leq \frac{X-\mu}{\sigma}\leq \frac{200-166.67}{11.79}]$

= $P[-1.41\leq Z\leq 2.83]=P[Z\leq 2.83]-P[Z$

= $\phi(2.83)-\phi(-1.41)$

=0.9977-0.0793=0.9184

Thus, the probability that the number 6 appears between 150 to 200 times=0.92

Now, given that 6 appears exactly 200 times .

Therefore, other number appear in other 800 rolls .

We have to find the probability that the number 5 will appear less than 150 times.

Therefore, for 800 rolls, let the event of 5 coming up be success.

Then , p=P(success)=P(5)= $\frac{1}{5}$

Let Y be the random variable denoting the number of times 5 coming up in 800 rolls.

Then, $Y\sim bin(800,\frac{1}{5})$

Mean, $\mu=np=800\times \frac{1}{5}=160$

Variance, $\sigma^2=np(1-p)=800\times \frac{1}{5}(1-\frac{1}{5})=128$

$Y\sim N(160,128)$ because n is large

$P(Y$

$P(Y$

Hence, the probability that the number 5 will appear less than 150 times given that 6 appeared exactly 200 times=0.19

4 0

2 years ago