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3 years ago

For 20 points and brainliest

Mathematics

2 answers:

erastova [34]3 years ago

5 0

Answer:

SA = 376.8 in²

Step-by-step explanation:

area of top:

π4² = 50.24 in²

50.24 x 2 = 100.48 in² (for both ends)

area of side:

circumference = 2 x 4 x π = 25.12 in

area = 25.12 x 11 = 276.32 in²

276.32 + 100.48 = 376.8 in²

Korvikt [17]3 years ago

4 0

Answer:

376.99in³

Step-by-step explanation:

no proof but a g00gle calculator, you're just gonna have to trust me on this one

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A job fair was held at the Student Union. 25% of the students who attended received job offers. Of all of the students at the jo

vovangra [49]

Answer:

A) Both events are not independent.

B) Both events are not mutually exclusive

C) 8.33%

D) 80%

Step-by-step explanation:

A) Both events are not independent. This is because, If B occurs it means that it is very likely that J will occur as well.

B) Both events are not mutually exclusive. This is because it is possible for both events J and B to occur at the same time.

C) we want to find the probability that Joe who is not a business student will receive the job offer.

This is;

P(J|Not B) = P(J & Not B)/P(Not B)

Now,

P(J & Not B) = P(J) – (P(B) × P(J | B))

25% of the students who attended received job offers. Thus; P(J) = 0.25

40% were from the College of Business. Thus;

P(B) = 0.4

Among the business students, 50% received job offers. Thus;

P(J|B) = 0.5

Thus;

P(J & Not B) = 0.25 - (0.4 × 0.5)

P(J & Not B) = 0.25 - 0.2

P(J & Not B) = 0.05

Since P(B) = 0.4

Then, P(Not B) = 1 - 0.4 = 0.6

Thus;

P(J|Not B) = 0.05/0.6

P(J|Not B) = 0.0833 = 8.33%

D) This probability is represented by;

P(B | J) = P(B & J)/P(J)

P(B & J) = (P(B) × P(J | B)) = (0.4 × 0.5) = 0.2

P(B | J) = 0.2/0.25

P(B | J) = 0.8 = 80%

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3 years ago

got everything right but my teacher said great and then said im missing the graph ? how do i do that ?

Diano4ka-milaya [45]

Ur supposed to have a graph paper and line the 12 and the 3y and 3x

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3 years ago

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Fudgin [204]

Answer:

27+r

Step-by-step explanation:

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3 years ago

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There was a bag of counters. 45 were large! For every large counters 3/5 were small! The green was 300% than yellow, there were

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Answer:

12 is the correct answer

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2 years ago

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

3 years ago