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3 years ago

Does the line passing through the points (3, 3) and (7, 11) intersect the line passing through the points (5, 8) and (13, 24

Mathematics

1 answer:

victus00 [196]3 years ago

7 0

Answer:

I understand why you are asking about this Ammaria, these type of questions take a lot of understanding of graphing. Please feel ok asking me questions :)

Step-by-step explanation:

we are given two points

P1 = (3,3) in the form (x1,y1)

P2 = (7,11) in the form (x2,y2)

slope = m

m = (y2-y1) / (x2-x1)

m = (11-3) / (7-3)

m = 8 / 4

m = 2

now use the slope along with one of the given points, either one will work, i'll use P1 , and put that into the point-slope formula.

y-3 = 2(x-3)

y -3 = 2x -6

y = 2x -3

So with the above formula we can check to see if the points work by plugging them into the formula.

P3 = (5,8)

8 = 2(5) - 3

no point 3 does not work so it's not on the line

P4 = (13, 24)

24 = 2(13) - 3

no point 4 does not work either.

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%