<span>Look at your table for a Z value of 1.55. The numbers on the far left column are your z values. See the 1.5 row, then move over to the 0.05 column to make it 1.55.
You'll see 0.9394.
That's the area under the normal curve from 1.55 to negative infinity.
But you wanted the area under the curve greater than 1.55.
Take 1-0.9394=0.0606.
You subtract from 1 because you know that the area under the whole curve is 1, so it gives you the area you need.</span>
Answer:
Step-by-step explanation:
mean is average, so if we label the first through fifth days of january a-e we can solve it.
So average of the first four is (a+b+c+d)/4 = 1 and the first five is (a+b+c+d+e)/5 = 2. Since it's easy, let's get rid of the fraction in both.
(a+b+c+d)/4 = 1 and (a+b+c+d+e)/5 = 2
a+b+c+d = 4 and a+b+c+d+e = 10
Now, we know a+b+c+d is 4, so we can replace that in the second equation
a+b+c+d+e = 10
4 + e = 10
e = 6
ad since e is the fifth day, we know the temperature of it. Let me know if something here didn't make sense.
B.
Because the input is time.
That narrows it down to B & C.
The time is in seconds. B.
Answer:
<u>Triangle</u>
base 8
Height 6
Area 48
<u>Rectangle</u>
Length 16
Width 10
Area 160
48+48+160=256
<u>Total Area of the Hexagon</u>
256
