For what values of k will the graph of y=kx+3 be a descending line?

As a fraction (all of the bellow) 6.7 -0.92 -0.1 -5.05

iren2701 [21]

Answer:

6.7 = 6 $\frac{7}{10}$

-0.92= - $\frac{23}{25}$

-0.1= - $\frac{1}{10}$

-5.05=-5 $\frac{1}{20}$

Step-by-step explanation:

:-)

First rewrite the decimal as a number with 1 in the denominator.

2nd multiply to remove 2 decimal places by (100/100)

3rd reduce the result of the multiplication

Then simplify the improper fraction

-Alex :-)

6 0

3 years ago

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The ages of three children in a family can be expressed as consecutive integers. The square of the age of the youngest child is

tensa zangetsu [6.8K]

Answer: 10, 11, & 12

Step-by-step explanation:

Let x represent the age of the youngest child.

Their ages are consecutive so,

Youngest: x

Middle: x + 1

Oldest: x + 2

The age of the Youngest squared (x²) equals 8 times the Oldest [8(x + 2)] plus 4.

x² = 8(x + 2) + 4

x² = 8x + 16 + 4

x² = 8x + 20

x² - 8x - 20 = 0

(x - 10)(x + 2) = 0

x - 10 = 0 or x + 2 = 0

x = 10 or x = -2

Since age cannot be negative, x = -2 is not valid

So, the Youngest (x) is 10

the Middle (x + 1) is 11

and the Oldest (x + 2) is 12

4 0

3 years ago

What is 30% of 70? Please help.

garik1379 [7]

Hi lovely,

The answer you're looking for is 21.

3 0

3 years ago

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What are all of the real roots of the following polynomial? f(x) = x4 - 13x2 + 36

GREYUIT [131]

ANSWER

A. -3, -2, 2, and 3

EXPLANATION

The given polynomial function is,

$f(x) = {x}^{4} - 13 {x}^{2} + 36$

To find the real roots, we equate the function to zero to obtain;

${x}^{4} - 13 {x}^{2} + 36 = 0$

We can solve this equation as a quadratic equation in
${x}^{2}.$

Thus we rewrite the equation as,

$({x}^{2})^{2} - 13 {x}^{2} + 36 = 0$

We split the middle term to get,

$({x}^{2})^{2} - 9 {x}^{2} - 4 {x}^{2} + 36 = 0.$

We factor to get,

${x}^{2} ( {x}^{2} - 9) - 4( {x}^{2} - 9) = 0$

We factor to get,

$( {x}^{2} - 9)( {x}^{2} - 4) = 0$

Either

${x}^{2} - 9 = 0$
or

${x}^{2} - 4 = 0$

$x = \pm \sqrt{9} \: or \: x = \pm \sqrt{4}$
$x = \pm3 \: or \: x = \pm2$

$x=-3,-2,2,3$

The correct answer is A

5 0

3 years ago

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Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-\frac{1}{1+\lambda}$

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

4 0

3 years ago