Answer:
-6.134 to +6.134
Step-by-step explanation:
given that a large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and a normal distribution
X is Normal with mean =6.1 and std dev = 1 unit
We are to determine the range of values containing 70% of the population of x
We know that normal distribution curve is bell shaped symmetrical about the mean.
So to find 70% range we can use 35% on either side of the mean
Using std normal distribution table the value of z for which probability from 0 to z is 0.35 is 1.034
Hence corresponding x value is

i.e. 70% values lie between
-6.134 to +6.134
Answer:
I could do 1 and 3
1) 2x-3y=-2 ....1
-
2x+y=14......2
=-4y=-16
y=4
<u>Substitute</u><u> </u><u>(</u><u>y</u><u>=</u><u>4</u><u>)</u><u> </u><u>into</u><u> </u><u>equation</u><u> </u><u>1</u>
2x-3 (4)=-2
2x-12=-2
2x=-2+12
2x=10
×=5
3) 5x+5y=20....1
-
-3x+5y=4......2
=8x=16
x=2
<u>S</u><u>ubstitute</u><u> </u><u>(</u><u>x</u><u>=</u><u>2</u><u>)</u><u> </u><u>into</u><u> </u><u>equation</u><u> </u><u>1</u>
<u>5</u><u> </u><u>(</u><u>2</u><u>)</u><u>+</u><u>5y</u><u>=</u><u>20</u>
<u>10</u><u>+</u><u>5y</u><u>=</u><u>20</u>
<u>5y</u><u>=</u><u>20-10</u>
<u>5y</u><u>=</u><u>10</u>
<u>y</u><u>=</u><u>2</u>
Answer:
m∠ABC = 130°
Step-by-step explanation:
BD is the angle bisector of angle ABC .
Therefore, ∠ABD ≅ ∠DBC
Measure of angle DBC = 65°
Therefore, m∠ABD = m∠DBC = 65°
m∠ABC = m∠ABD + m∠DBC
= 65° + 65°
= 130°