A brochure claims that the average maximum height for a certain type of plant is 0.7 m. A gardener suspects that this is not accurate locally due to variation in soil conditions, and believes the local height is shorter. A random sample of 40 mature plants is taken. The mean height of the sample is 0.65 m with a standard deviation of 0.20 m. Test the claim that the local mean height is less than 0.7 m using a 5% level of significance.

As , it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.

Step-by-step explanation:

1. Relevant data:

2. Hypotesis testing

3. Find the rejection area

From the one tail standard normal chart, whe have Z-value for is 1.56

Then rejection area is left 1.56 in normal curve.

4. Find the test statistic:

5. Hypotesis Testing

As , it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

⭐⭐

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3