Help ASAP!!! 19 points!!!

See the picture attached to better understand the problem

we know that
If two secant segments are drawn to a circle from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
so
jl*jk=jn*jm------> jn=jl*jk/jm

we have
jk=8,lk=4 and jm=6
jl=8+4----> 12

jn=jl*jk/jm-----> jn=12*8/6----> jn=16

the answer is
jn=16

6 0

3 years ago

Using traditional methods, it takes 10.1 hours to receive a basic driving license. A new license training method using Computer

Law Incorporation [45]

Answer:

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis: $\mu =10.1$

Alternative hypothesis: $\mu \neq 10.1$

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$

$p_v =2*P(t_{(11)}>0.990)=0.343$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=10.5$ represent the sample mean

$s=1.4$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =10.1$ represent the value that we want to test

$\alpha=0.1$ 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.

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis: $\mu =10.1$

Alternative hypothesis: $\mu \neq 10.1$

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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".

Calculate the statistic

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

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=12-1=11$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(11)}>0.990)=0.343$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.

7 0

3 years ago