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lyudmila [28]
3 years ago
11

How can you express (15 30) as a multiple of a sum of whole numbers with no common factor?

Mathematics
2 answers:
tigry1 [53]3 years ago
7 0

Answer with explanation:

We have to write ,15 and 30 as a multiple of sum of whole numbers with no common factor.

→15= 13+2----13 and 2 are both prime that is no factor other than 1 and itself.

→30=13 +17---13 and 17 are both prime that is no factor other than 1 and itself.

Alex_Xolod [135]3 years ago
4 0
Well theoretically speaking you could but that would take a long time and 15 and 30 have a common factor of 5 and 15. So I think for this one you can't since there are two common factors.
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If r and s are positive integers, is \small \frac{r}{s} an integer? (1) Every factor of s is also a factor of r. (2) Every prime
Yuri [45]

Answer:

<em>If statement(1) holds true, it is correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em>If statement(2) holds true, it is not necessarily correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em></em>

Step-by-step explanation:

Given two positive integers r and s.

To check whether \small \frac{r}{s} is an integer:

Condition (1):

Every factor of s is also a factor of r.

r \geq s

Let us consider an example:

s = 5^2 \cdot 2\\r = 5^3 \cdot 2^2

\dfrac{r}{s} = \dfrac{5^3\cdot2^2}{5^2\cdot2} = 10

which is an integer.

Actually, in this situation s is a factor of r.

Condition 2:

Every prime factor of <em>s</em> is also a prime factor of <em>r</em>.

(But the powers of prime factors need not be equal as we are not given the conditions related to powers of prime factors.)

Let

r = 2^2\cdot 5\\s =2^4\cdot 5

\dfrac{r}{s} = \dfrac{2^3\cdot5}{2^4\cdot5} = \dfrac{1}{2}

which is not an integer.

So, the answer is:

<em>If statement(1) holds true, it is correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em>If statement(2) holds true, it is not necessarily correct that </em>\small \frac{r}{s}<em> is an integer.</em>

<em></em>

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3 years ago
Brainliest* and show work and the answer to number 15
barxatty [35]

Step-by-step explanation:

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2 years ago
Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p
Ymorist [56]

Answer:

E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

A\sim Uniform(1 ,7)

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

B\sim Exp(\lambda=\frac{1}{30 min})

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

E(A +B)

Since the two random variables are assumed independent, then we have this

E(A+B) = E(A)+E(B)

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by E(X) =\frac{a+b}{2} where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

E(A)=\frac{1+7}{2}=4 hours

The expected value for the exponential distirbution is given by :

E(X)= \int_{0}^\infty x \lambda e^{-\lambda x} dx

If we use the substitution y=\lambda x we have this:

E(X)=\frac{1}{\lambda} \int_{0}^\infty y e^{-\lambda y} dy =\frac{1}{\lambda}

Where X represent the random variable and \lambda the parameter. If we apply this formula to our case we got:

E(B) =\frac{1}{\lambda}=\frac{1}{\frac{1}{30}}=30min

We can convert this into hours and we got E(B) =0.5 hours, and then we can find:

E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours

And in order to find the variance for the random variable A+B we can find the individual variances:

Var(A)= \frac{(b-a)^2}{12}=\frac{(7-1)^2}{12}=3 hours^2

Var(B) =\frac{1}{\lambda^2}=\frac{1}{(\frac{1}{30})^2}=900 min^2 x\frac{1hr^2}{3600 min^2}=0.25 hours^2

We have the following property:

Var(X+Y)= Var(X)+Var(Y) +2 Cov(X,Y)

Since we have independnet variable the Cov(A,B)=0, so then:

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

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Answer:

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Step-by-step explanation:

Let the payment for regular hours be $x

given that

Brandon is paid 150% of his regular hourly rate for overtime hours

payment for overtime hours = 150% of payment for regular hours

payment for overtime hours = 150/100 * x = 3x/2

Given that He is paid \$45.00 an hour for overtime hours

thus,

3x/2 = 45

=> x = 45*2/3 = 30

Thus,  regular hourly rate for Brandon is $30

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