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valentinak56 [21]
3 years ago
6

For each sentence below, find the value of x that makes each sentence true.

Mathematics
1 answer:
vitfil [10]3 years ago
4 0

<u>ANSWER</u>

1. x=\frac{1}{2}


2. x=1


<u>QUESTION 1</u>

The first sentence is (5^{\frac{1}{5}})^5=25^x.


Recall that;

(a^m)^n=a^{mn}


We simplify the left hand side by applying this property to get;


5^{\frac{1}{5}\times 5}=25^x.


\Rightarrow 5^{1}=25^x.


We now rewrite the right hand side too in an index form to obtain;


\Rightarrow 5^{1}=5^{2x}

We now equate the exponents to get;

\Rightarrow 1=2x.


\Rightarrow \frac{1}{2}=x


\Rightarrow x=\frac{1}{2}.


<u>QUESTION 2</u>

The second sentence is (8^{\frac{1}{3}})^2=4^x


We simplify the left hand side first to get;

(2^{3\times \frac{1}{3}})^2=4^x


2^2=4^x


We now rewrite the left hand side too in index form to obtain;


2^2=2^{2x}

We equate the exponents to get;

2=2x


This implies that;


1=x

or

x=1

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Answer:

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(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> =  Mike will show up.

Given:

P(D^{c})=0.55\\P(M^{c})=0.45

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

                                                   =1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25

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There are 21 ways in which the 5-committee member can be selected from the 7 volunteers.
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