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3 years ago

What is the equation in point-slope form of the line passing through (1, 9) and (−1, 11)?

1 answer:

7 0

Point slope form:
$y-y_1=m(x-x_1)$

m=slope
(x1,y1)

First we need to find the slope using the slope formula:
$\frac{y_2-y_1}{x_2-x_1} =slope$

Given the two coordinates:
(1, 9) and (−1, 11)

Plug them into the slope formula:
 $\frac{11-9=2}{-1-1=-2} =slope$
Simplify:
$\frac{2}{-2} =-1$

Now we know that:
m=-1 and the (x1,y1) would be (1, 9)

Now we simply plug in the info:
$y-9=-(x-1)$

Final answer: y − 9 = −(x − 1)

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Assume the random variable X is normally distributed with meanmu equals 50μ=50and standard deviationsigma equals 7σ=7.Compute th

nekit [7.7K]

Answer:

$P(35$

And we can find this probability with the following difference and using the normal standard table or excel we have:

$P(-2.143$

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".

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:

$X \sim N(50,7)$

Where $\mu=50$ and $\sigma=7$

We are interested on this probability

$P(35$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(35$

And we can find this probability with the following difference and using the normal standard table or excel we have:

$P(-2.143$

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3 years ago

In a casino game, gamblers are allowed to roll n fair, 6-sided dice. If a 6 shows up on any of them, the gambler gets nothing. I

Soloha48 [4]

Answer:

a) $\bf \boxed{W(n) = 3*n*\left( \frac{5}{6} \right)^n\;dollars}$

the smallest n that maximizes the expected payoff is n=5.

b) 4

Step-by-step explanation:

The expected amount of $ won for each die would be the average of 1, 2, 3, 4 and 5 which is $3.

Let W(n) the expected money won when rolling n dice.

n =1

If the gambler rolls only one die, the expected money won would be $3 times the probability of not getting a 6, which is 5/6.

$\bf W(1) = 3*1*\frac{5}{6}$

n=2

If the gambler rolls 2 dice, the expected money won would be $3 times the probability of not getting a 6 in any of the dice. Since the outcome of the rolling does not depend on the previous rollings, the probability is

$\bf \frac{5}{6}\times\frac{5}{6}=\left( \frac{5}{6} \right)^2$

and

$\bf W(2) = 3*2*\left( \frac{5}{6} \right)^2$

n=3

Similarly, since the probability of not getting a 6 in 3 dice equals

$\bf \left( \frac{5}{6} \right)^3$

$\bf W(3) = 3*3*\left( \frac{5}{6} \right)^3$

and the formula for the expected money won with n dice would be

$\bf \boxed{W(n) = 3*n*\left( \frac{5}{6} \right)^n\;dollars}$

In the picture attached there is a plot of the values of the expected money won for n=1 to 20 (See picture)

For n=5 and n=6 we get the maximum profit expected of $6.02816=$6 rounded to the nearest integer.

Hence, the smallest n that maximizes the expected payoff is n=5.

The probability that face k (k=1,2,...or 6) shows up is 1/6,

as this face can be in any of the 10 positions of the arrangement, there are 10 ways that face k can show up.

The probability that face k (k=1,2,...or 6) shows up twice is $\bf \left( \frac{1}{6} \right)^2$

as this face can be in any of the $\bf C(10;2)=\binom{10}{2}$ (combinations of 10 taken 2 at a time) positions of the arrangement, there are C(10;2) ways that face k can show up twice.

The probability that face k (k=1,2,...or 6) shows up three times is $\bf \left( \frac{1}{6} \right)^3$

as this face can be in any of the $\bf C(10;3)=\binom{10}{2}$ (combinations of 10 taken 2 at a time) positions of the arrangement, there are C(10;3) ways that face k can show up twice.

So, we infer that the expectation is

$\bf \sum_{k=1}^{10}\binom{10}{k}(1/6)^k=3.6716\approx 4$

and the expected number of distinct dice values that show up is 4.

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3 years ago