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3 years ago

The point (5,-6) lies on a circle and the center of the circle is at the point (5,-8).

Mathematics

1 answer:

mario62 [17]3 years ago

3 0

Answer:

(x - 5)^2 + (y + 8)^2 = 4

Step-by-step explanation:

The equation for a circle in center radius form is

(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius

We need to determine the radius, from the two points which is the distance

d = sqrt( (x2-x1)^2 + (y2-y1)^2)

= sqrt( (5-5)^2+( -8--6)^2)

= sqrt(0+ (-8+6)^2)

= sqrt( 0+(-2)^2)

= sqrt(4)

= 2

The radius is 2 and the center is (5,-8)

(x-5)^2 + (y--8)^2 = 2^2

(x-5)^2 + (y+8)^2 = 4

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If a coin is tossed 5 times, and then a standard six-sided die is rolled 4 times, and finally a group of three cards are drawn f

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Answer: 5,499,187,200

<u>Step-by-step explanation:</u>

A coin is tossed 5 times.

There are two options (heads or tails) so the possible outcomes are: 2⁵

A six-sided die is rolled 4 times.

There are six options so the possible outcomes are: 6⁴

A group of 3 cards are drawn (without replacement).

The first outcome has 52 options, the second has 51 options, and the third has 50 options: 52 x 51 x 50

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3 years ago

A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe

d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

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3 years ago

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lesantik [10]

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