Line GH contains points G(–2, 6) and H(5, –3). Given the points, what is the slope of the line ?.

2 answers:

6 0

The slope is m = rise/run
m = ( -3 - 6 ) / ( 5 - ( -2 )) = -9 / ( 5 + 2 ) = - 9/7
Answer: the slope of the line is -9/7.

user100 [1]3 years ago

4 0

The term to find the slope is Y2-Y1/X2-X2. So it becomes -3-6/5-(-2) which translates to -3-6/5+2 and YOUR FINAL ANSWER IS: -9/7.

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In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

$\mu_s=np=500*0.90=450$

The standard deviation will be:

$\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7$

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

$z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932$

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

$\mu=np=300*0.932=279.6$

The standard deviation will be:

$\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36$

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

$P(270\leq x\leq280)=P(269.5$

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

$P(X=280)=P(279.5$

8 0

3 years ago

Determine if the following lengths could represent a triangle. If they can, determine if the triangle, is acute, obtuse, or righ

stiks02 [169]

9514 1404 393

Answer:

A. no

B. no

C. obtuse

Step-by-step explanation:

For side lengths to form a triangle, the sum of the shorter two must exceed the longest.

A. 5 + 8 = 13 . . . . a line segment, not a triangle

B. 7 + 12 < 26 . . . . no closure, not a triangle

C. 11 + 15 > 20 . . . . a triangle. A picture shows it to be obtuse

You can also compare 11² +15² vs 20² ⇒ 346 vs 400. The long side is too long for a right triangle, so the triangle must be obtuse. (The Pythagorean theorem tells you a right triangle with those legs would have a long side of √346 = 18.6.)