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BaLLatris [955]
2 years ago
5

Find the probability.

Mathematics
1 answer:
pogonyaev2 years ago
5 0

|\Omega|=14\cdot13=182\\|A|=6\cdot4=24\\\\P(A)=\dfrac{24}{182}=\dfrac{12}{91}\approx13.2\%

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Gravel is being dumped from a conveyor belt at a rate of 15 ft3/min, and its coarseness is such that it forms a pile in the shap
gayaneshka [121]

Answer:

0.13 ft/min

Step-by-step explanation:

We are given that

\frac{dV}{dt}=15ft^3/min

We have to find the increasing rate of change of height of pile  when the pile is 12 ft high.

Let d be the diameter of pile

Height of pile=h

d=h

Radius of pile,r=\frac{d}{2}=\frac{h}{2}

Volume of pile=\frac{1}{3}\pi r^2 h=\frac{1}{12}\pi h^3

\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}

h=12 ft

Substitute the values

15=\frac{1}{4}\pi(12)^2\frac{dh}{dt}

\frac{dh}{dt}=\frac{15\times 4}{\pi(12)^2}

\frac{dh}{dt}=0.13ft/min

7 0
3 years ago
What is the benchmark of 6/10
IceJOKER [234]
First split your model into 10 equal sections then you shade/color in 6 of them.
There it goes that is 6/10 bench mark!
~JZ
Hope it helps
3 0
3 years ago
The ratio of credit card sales to cash sales at a store was 8:3. If the store had 144 cash sales, how many more credit card sale
Neko [114]

Answer:

There were 240 more credit card sales than cash sales.

Step-by-step explanation:

By using the ratio 8:3 I was able to determine that if there were 144 cash sales that would mean that there would be 384 credit card sales. However, the question asks "How many <u>more</u> credit card sales were there than cash sales?" Meaning that we need to find the difference. 384-144= 240.

This means that there were 240 <u>more</u> credit card sales than cash sales.

Hope this helped! If you have any questions about what I did please comment!

And if this helped please mark brainliest!

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4 0
3 years ago
The rectangle below has an area of 70y⁸ 30y⁶. The width of the rectangle is equal to the greatest common monomial factor of 70y⁸
Lorico [155]

Answer:

The Width of the Rectangle =10y^6

\text{Length of the Rectangle}=210y^8

Step-by-step explanation:

The area of the rectangle =70y^8 30y^6.

We are told that the width of the rectangle is equal to the greatest common monomial factor of 70y^8 \: and\: 30y^6.

Let us determine the greatest common monomial factor of 70y^8 \: and\: 30y^6.

Express each term as a product to pick out the common factors:

70y^8 =7X10Xy^6Xy^2\\30y^6=3X10Xy^6

In the two terms, the common terms are 10 and y^6. Therefore their greatest monomial factor =10y^6

The Width of the Rectangle =10y^6

Recall: Area of a Rectangle =Length X Width

70y^8 30y^6=Length X 10y^6\\Length =70y^8 30y^6 \div 10y^6 \\=\dfrac{70X30Xy^8Xy^6}{10y^6} =210y^8\\\text{Length of the Rectangle}=210y^8

6 0
3 years ago
Read 2 more answers
How many seconds will light leaving Los Angeles take to reach the following locations (a) San Francisco (about 500km), (b) Londo
kondor19780726 [428]

Answer:

a) It takes 0.0017s for the light to reach San Francisco.

b) It takes 0.033s for the light to reach London.

c) It takes 1.334s for the light to reach Mars.

d) It takes 149.7s for the light to reach Venus.

Step-by-step explanation:

Here we can solve this problem by using this following formula:

s = \frac{d}{t}

In which s is the speed(in km/s), d is the distance(in km) and t is the time(in s).

The light speed is 299 792 458 m / s = 299,792.458 km/s, so s = 299,792.458

(a) San Francisco (about 500km)

Find t when d = 500. So

s = \frac{d}{t}

299,792.458 = \frac{500}{t}

299,792.458t = 500

t = \frac{500}{299,792.458}

t = 0.0017s

It takes 0.0017s for the light to reach San Francisco.

b) London(about 10,000km)

Find t when d = 10,000. So

s = \frac{d}{t}

299,792.458 = \frac{10,000}{t}

299,792.458t = 10,000

t = \frac{10,000}{299,792.458}

t = 0.033s

It takes 0.033s for the light to reach London.

(c) the Moon (400,000km)

Find t when d = 400,000. So

s = \frac{d}{t}

299,792.458 = \frac{400,000}{t}

299,792.458t = 400,000

t = \frac{400,000}{299,792.458}

t = 1.334s

It takes 1.334s for the light to reach Mars.

(d) Venus (0.3 A.U. from Earth at its closest approach).

Each A.U. has 149,59,7871 km.

So 0.3 A.U. = 0.3*(149,597,871) = 44,879,361.3. This means that d = 44,879,361.3. So:

s = \frac{d}{t}

299,792.458 = \frac{44,879,361.3}{t}

299,792.458t = 44,879,361.3

t = \frac{44,879,361.3}{299,792.458}

t = 149.7s

It takes 149.7s for the light to reach Venus.

3 0
3 years ago
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