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Rudiy27
1 year ago
7

After being filled, a basketball loses 3.2% of its air every day. Considering that the initial volume of the ball was 840 cubic

centimetres when it was inflated, what will be the volume in cubic centimetres after 4 days? (Round your answer to one decimal place).
Mathematics
2 answers:
Sliva [168]1 year ago
8 0

Answer:

volume = 840 cubic cm

(3.2/100)×840= 26.88 per day

after 4 days = 26.88×4=107.5

answer is 107.5

Brilliant_brown [7]1 year ago
4 0

Answer:

Step-by-step explanation:

V(d)= the total amount of air in the ball

V(4)=840(1-0.032)^4 = 840(0.968)^4= 737.53cm^3 which to one decimal place is 737.5cm^3

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Answer:

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Step-by-step explanation:

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3 years ago
Hii I need help with these questions tyy!
9966 [12]

Answer:

(7x+2)(3x-1); (5x+4)^2 or (5x+4)(5x+4);

Step-by-step explanation:

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21.

using the formula to find area...

l*w=a

(7x+2)(3x-1) would be your answer

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22.

since the problem states that it's a square, you can just say either...

(5x+4)^2 or (5x+4)(5x+4) not sure which one your teacher accepts

4 0
2 years ago
Read 2 more answers
6(2 + 4) -1 divided by 2.3+1
Tcecarenko [31]

Answer:

when 6(2+4) -1 divided by 2.3 + 1 the answer is 36.5652174

6 0
3 years ago
A warehouse worker fills 150 orders per day on average. From day to day, the number of orders varies by 2%. What is the range of
choli [55]

Answer:

The range is from 147 to 153 orders per day

Step-by-step explanation:

orders varies by 2% means that orders can be LOWER THAN THE AVERAGE, or HIGHER THAN THE AVERAGE.

That is, by 2%.

First, we need to find the decimal of 2%, so

2/100 = 0.02

We multiply this with the average number of order, 150, to get the varying amount:

0.02 * 150 = 3

Thus, the range would be:

150 - 3 = 147

150 + 3 = 153

The range is from 147 to 153 orders per day

5 0
3 years ago
Please answer all parts of the question and all work shown.
faust18 [17]

Answer:

a. 0.4931

b. 0.2695

Step-by-step explanation:

Given

Let BG represents Boston Globe

NYT represents New York Times

P(BG) = 0.55

P(BG') = 1 - 0.55 = 0.45

P(NYT) = 0.6

P(NYT') = 1 -0.6 = 0.4

Number of headlines = 5

Number of depressed articles = 3 (at most)

a.

Let P(Read) = Probability that he reads the news the first day

P(Read) = P(He reads BG) and P(He reads NYT)

For the professor to read BG, then there must be at most 3 depressing news

i.e P(0) + P(1) + P(2) + P(3)

But P(0) + P(1) + .... + P(5) = 1 (this is the sample space)

So,

P(0) + P(1) + P(2) + P(3) = 1 - P(4) - P(5)

P(4) or P(BG = 4) is given as the binomial below

(BG + BG')^n where n = 5, r = 4

So, P(BG = 4) = C(5,4) * 0.55⁴ * 0.45¹

P(BG = 5). = (BG + BG')^n where n = 5, r = 5

So, P(BG = 5) = C(5,5) * 0.55^5 * 0.45°

P(0) + P(1) + P(2) + P(3)= 1 - P(BG = 4) - P(BG = 5)

P(0) + P(1) + P(2) + P(3) = 1 - C(5,4) * 0.55⁴ * 0.45¹ - C(5,5) * 0.55^5 * 0.45°

P(0) + P(1) + P(2) + P(3) = 0.7438

For the professor to read NYT, then there must be at most 3 depressing news

i.e P(0) + P(1) + P(2) + P(3)

But P(0) + P(1) + .... + P(5) = 1 (this is the sample space)

So,

P(0) + P(1) + P(2) + P(3) = 1 - P(4) - P(5)

P(4) or P(NYT = 4) is given as the binomial below

(NYT+ NYT')^n where n = 5, r = 4

So, P(NYT = 4) = C(5,4) * 0.6⁴ * 0.4¹

P(NYT = 5). = (NYT + NYT')^n where n = 5, r = 5

So, P(NYT = 5) = C(5,5) * 0.6^5 * 0.4°

P(0) + P(1) + P(2) + P(3)= 1 - P(NYT = 4) - P(NYT = 5)

P(0) + P(1) + P(2) + P(3) = 1 - C(5,4) * 0.6⁴ * 0.4¹ - C(5,5) * 0.6^5 * 0.4°

P(0) + P(1) + P(2) + P(3) = 0.6630

P(Read) = P(He reads BG) and P(He reads NYT)

P(Read) = 0.7438 * 0.6630

P(Read) = 0.4931

b.

Given

n = Number of week = 7

P(Read) = 0.4931

R(Read') = 1 - 0.4931 =

He needs to read at least half the time means he reads for 4 days a week

So,

P(Well-informed) = (Read + Read')^n where n = 7, r = 4

P(Well-informed) = C(7,4) * (0.4931)⁴ * (1-0.4931)³ = 0.2695

3 0
3 years ago
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