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3 years ago

The diagram shows a wooden prism of height 5cm.

Mathematics

1 answer:

igomit [66]3 years ago

5 0

Answer:

280.8997

Step-by-step explanation:

cross section = 2*15*5+2*(25/360)*15*15*pi+5*(25/360)*2*pi*15

= 280.8997

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Write the value of nine 913, 256

monitta

It is the hundredth thousandth (: If that is what you mean.

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3 years ago

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Evaluate x = -3.75 4x + 12 - 8x + 23 =

yanalaym [24]

Answer:

u have Isolate the variable by dividing each side by factors that don't contain the variable. and the correct answer is x=2.7442371

Step-by-step explanation:

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3 years ago

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A doctor orders 150 mg of aspirin to be taken by 12.4 Ib infant every 9 hours. How many mg of the drug does the infant take in 3

andreev551 [17]

Answer: 600 mg in 36 hours.

Step-by-step explanation:

1. Divide 36 by 9, so you know how many times the infant has to take the medicine. 36/9=4

2. Multiply 150x4 because the infant has to take 150 mg, each dosage. 150x4=600.

3. Your final answer is 600 mg. (You did not need the infant's weight in this problem. So, that's why it was not worked in.)

Hope this helps. :)

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4 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

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3 years ago

What is the value of 1: 7³ 2: 2⁶ 3: 4⁴

Veseljchak [2.6K]

1) 343
7 • 7 • 7 = 343

2) 64
2 • 2 • 2 • 2 • 2 • 2 = 64

3) 256
4 • 4 • 4 • 4 = 256

7 0

3 years ago

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