#CarryOnLearning
You ride a bicycle at a speed of 18 miles per hour. I am 3 miles behind my friend. My friend rides a bicycle at a speed of 12 mi
1 answer:
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As your testing H0:p=0.79, then you have a two-tailed test.
The p-value at two-tailed test is given by:
Then, you need to check into the standard normal cumulative table to find at which -z you do have a probability of 0.395. Thus:
As you can see in the picture at z=-0.26 you have a P(Z<=-z)=0.3974
And at z=-0.27 you have a P(Z<=-z)=0.3936.
The average of these values is:
Then, you have a probability of 0.395 at a z of:
Then -z=-0.27, thus z=0.27
Answer:
1)
Step-by-step explanation:
Let's assume we know the measurement for angle F (let's just call it 20 degrees to help us imagine things).
The triangle in the right of the drawing and the bigger triangle (that contains both of the triangles) share that angle.
We are given angleEDF has measurement 90 degrees because of the square there.
We can find angleE's measurement using that the sum of the interior angles of a triangle is 180.
I'm looking at the green triangle you have 90+20+mAngleE=180.
Let's solve that for mAngleE.
90+20+mAngleE=180
110+mAngleE=180
Subtract 110 on both sides:
mAngleE=180-110
mAngle=70.
The only way we are going to prove these triangles are similar is if we have a condition that helps us show that the smaller two triangles are also 20-70-90 triangles.
Let's look at choice 1)
If DG is an altitude then DG is perpendicular to EF. Perpendicular means it forms 90 degree angles at the intersection.
If we look at the right triangle, this gives us enough information to find the last angle, the angle known as GDF.
mAngleGDF+20+90=180
mAngleGDF+110=180
Subtract 110 on both sides:
mAngleGDF=70.
So we have a 20-70-90 triangle again. So the one on the right and the green triangle are similar.
We can find angle EDG's measurement to since we know angle E's measurement is 70 and angle EGD's measurement is 90.
mAngleEDG+70+90=180
mAngleEDG+160=180
mAngleEDG=20
So we have a 20-70-90 triangle again. So the one on the left is similar to the other two triangles in this picture as well.
Again if the corresponding angles of a triangle are congruent then the triangles are at least similar.
I provided a drawing for number 1 with the example I put here.
Let's look at 2). We already know EF is an hypotenuse of triangle EDF since it is opposite the 90 angle that is EDF.
2 doesn't help us because it gives no additional information.
Let's look at 3). Segment ED is shorter than segment DF. This gives us no measurements or even a ratio to compare the sides.
Let's look at 4). Angle is congruent to itself. There is only one angle that can be seen as angle E in this picture so this also provides no additional information.
