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KATRIN_1 [288]
3 years ago
8

Expanding Logarithmic Expressions In Exercise, use the properties of logarithms to rewrite the expression as a sum, difference,

or multiple of logarithms.
In (x^2(x - 1))1/2
Mathematics
1 answer:
aev [14]3 years ago
6 0

Answer:

The simplified expression is:

\ln{x} + \frac{\ln{(x-1)}}{2}

Step-by-step explanation:

We have those following logarithmic properties:

\ln{\frac{a}{b}} = \ln{a} - \ln{b}

\ln{a*b} = \ln{a} + \ln{b}

\ln{a^{n}} = n\ln{a}

In this problem, we have that:

\ln{x^{2}(x-1)}^{1/2}

Applying these properties

\frac{1}{2}*(\ln{x^{2} + \ln{(x-1)})

\ln{x} + \frac{\ln{(x-1)}}{2}

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13/15 + 1/8 =<br><br> Please help
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Reduce the fraction using 3
13/15+1/8

13/15=1/5

1/5+1/8

8+5/40

fraction response: 13/40
answer in decimal numbers: 0.325




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2 years ago
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Add 2/5 plus 27/9 -0
Leviafan [203]
2/5 plus 27/9 minus 0 equals 0
4 0
3 years ago
Find the value of x.
aleksley [76]

Answer:

38+82 =70

180-70=110

x=110°

5 0
3 years ago
34​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573

P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0
3 years ago
ASAP! Help! I’ll mark you brainly!
Alenkinab [10]

Answer:

y=45x+150

its a function but not a linear function

Step-by-step explanation:

hope this helps

4 0
3 years ago
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