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Leokris [45]
2 years ago
15

Christine performed the following experiment 20 times using a book that contains 462 pages.

Mathematics
1 answer:
wel2 years ago
7 0

Using the probability concept, it is found that there is a 0.3 = 30% experimental probability that Christine will randomly choose a page that is divisible by 3.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In an experimental probability, these number of outcomes are taken from previous trials.

In this problem:

  • The experiment was made 20 times.
  • Of those, in 6 experiments(81, 105, 171, 159, 297, 387) she choose a page that was divisible by 3.

Then:

p = \frac{6}{20} = 0.3

0.3 = 30% experimental probability that Christine will randomly choose a page that is divisible by 3.

You can learn more about the probability concept at brainly.com/question/15536019

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What is the answer to 5x+9x=100?
alexira [117]

Answer:

50/7.

Simplified to 7.14.

Step-by-step explanation:

Combine like terms, 5x and 9x.

5x+9x=14x

14x = 100

14x / 14 = 100 / 14

x = 50/7 or 7.14

Hope this helps!

5 0
2 years ago
Read 2 more answers
Which number is a solution of a² +4= 6a – 1?<br> 07<br> 6<br> 7<br> 8
LuckyWell [14K]

Answer:

a=1 or a=5

Step-by-step explanation:

a^{2}+4=6a−1

Step 1: Subtract 6a-1 from both sides.

a^{2}+4−(6a−1)=6a−1−(6a−1)

a^{2}−6a+5=0

Step 2: Factor left side of equation.

(a−1)(a−5)=0

Step 3: Set factors equal to 0.

a−1=0 or a−5=0

a=1 or a=5

(is it helpful rate me according to that

Thank You:-)

6 0
3 years ago
Please help with answers A and B :)
Over [174]
A. Terms
b. 9
64 to 49 is -15, 49 to 36 is -13, 36 to 25 is -11, 25 to 16 is -9, so the next number is 16-7=9
5 0
2 years ago
How do you solve x - 1 + 5x = 23​
Roman55 [17]

Answer:

X=4

Step-by-step explanation:

First you add like terms, in this case 5x and x. Then you have 6x-1=23. You then add 1 from both sides so you end up with 6x=24. This then results in X=4

8 0
2 years ago
Read 2 more answers
Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de
Ganezh [65]

Answer:

a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

The initial value problem is given as:

y' +y = 7+\delta (t-3) \\ \\ y(0)=0

Applying  laplace transformation on the expression y' +y = 7+\delta (t-3)

to get  L[{y+y'} ]= L[{7 + \delta (t-3)}]

l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

3 0
3 years ago
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