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coldgirl [10]
2 years ago
12

Josef and Timothy play a game in which Josef picks an integer between 1 and 1000 inclusive and Timothy divides 1000 by that inte

ger and states whether or not the quotient is an integer. How many integers could Josef pick such that Timothy's quotient is an integer
Mathematics
1 answer:
Reil [10]2 years ago
3 0
Timothy had played the game to often between 1 and 1000 but then divides that integer but his quotient is not integer the amount about the game he played
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Verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial 4x^3+20x^2+2x-3. Also verify the relationship between the ze
Finger [1]

Solution:

we have been asked to verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial 4x^3+20x^2+2x-3

To verify that whether the given values are zeros or not we will substitute the values in the given Polynomial, if it will returns zero, it mean that value is Zero of the polynomial. But if it return any thing other than zeros it mean that value is not the zero of the polynomial.

Let f(x)=4x^3+20x^2+2x-3\\\\\text{when x=-5}\\\\f(-5)=4(-5)^3+20(-5)^2+2(-5)-3=-13\\\\\text{when x=}\frac{1}{2}\\

f( \frac{1}{2} ) = 4 ( \frac{1}{2} )^3+20(\frac{1}{2})^2+2(\frac{1}{2})-3=\frac{7}{2}\\\\

\text{when x=}\frac{3}{4}\\\\

f( \frac{3}{4} ) = 4 ( \frac{3}{4} )^3+20(\frac{3}{4})^2+2(\frac{3}{4})-3=\frac{183}{16}\\

Hence -5, 1/2, and 3/4 are not the zeroes of the given Polynomial.

Since sum of roots=\frac{-b}{a}= \frac{-20}{4}=-5\\

But -5+\frac{1}{2}+\frac{3}{4}=  \frac{-15}{4}\neq-5

Hence we do not find any relation between the coefficients and zeros.

Anyway if the given values doesn't represents the zeros then those given values will not have any relation with the coefficients of the p[polynomial.

6 0
3 years ago
8 x - 4 (5 - x) = -44
Sphinxa [80]

8 x - 4 (5 - x) = -44

mutiply the bracket by -4

(-4)(5) = -20

(-4)(-x)= 4x

8x-20+4x= -44

8x+4x-20= -44 ( combine like terms )

12x-20= -44

move -20 to the other side

sign changes from -20 to +20

12x-20+20= -44+20

12x= -44+20

12x= -24

divide both sides by 12

12x/12= -24/12

Answer: x= -2

3 0
3 years ago
Stuck on two. Plz help. Pre-all is hard
algol [13]
Gncgbvcrvbshj hnctvvdf
7 0
3 years ago
سر
densk [106]

Answer:

3cm

Step-by-step explanation:

A particular satellite is 15 m wide

Model of it was built with a scale of 1 cm: 5 m

=> scale of the model will be: 1/500cm and  A particular satellite is 1500 cm wide

=>1500*1/500=3(cm)

7 0
3 years ago
A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.
AVprozaik [17]

Answer:

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean  is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\mu_{\bar X} = 61

\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715

So then is appropiate use the normal distribution to find the probabilities for \bar X

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  The letter \phi(b) is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: \phi(b)=P(z

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(61,14)  

Where \mu=61 and \sigma=14

Since the distribution for X is normal then the distribution for the sample mean \bar X is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\mu_{\bar X} = 61

\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715

So then is appropiate use the normal distribution to find the probabilities for \bar X

8 0
3 years ago
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