Answer:
Step-by-step explanation:
<u>The gray triangles have</u>
- Base of 20 cm and added height of 60 cm for 2 triangles in each section of AMND and MNCB
<u>The grey area in total is</u>
- 1/2*20*60 + 1/2*20*60 = 20*60 = 1200 cm²
Answer:
20%×X=30
x=150
Step-by-step explanation:
30÷20%=150
20%×150=30
How To Solve Systems of Inequalities Graphically
1) Write the inequality in slope-intercept form or in the form
y
=
m
x
+
b
y=mx+b
.
For example, if asked to solve
x
+
y
≤
10
x+y≤10
, we first re-write as
y
≤
−
x
+
10
y≤−x+10
.
2) Temporarily exchange the given inequality symbol (in this case
≤
≤
) for just equal symbol. In doing so, you can treat the inequality like an equation. BUT DO NOT forget to replace the equal symbol with the original inequality symbol at the END of the problem!
So,
y
≤
−
x
+
10
y≤−x+10
becomes
y
=
−
x
+
10
y=−x+10
for the moment.
3) Graph the line found in step 2. This will form the "boundary" of the inequality -- on one side of the line the condition will be true, on the other side it will not. Review how to graph a line here.
4) Revisit the inequality we found before as
y
≤
−
x
+
10
y≤−x+10
. Notice that it is true when y is less than or equal to. In step 3 we plotted the line (the equal-to case), so now we need to account for the less-than case. Since y is less than a particular value on the low-side of the axis, we will shade the region below the line to indicate that the inequality is true for all points below the line:
5) Verify. Plug in a point not on the line, like (0,0). Verify that the inequality holds. In this case, that means
0
≤
−
0
+
10
0≤−0+10
, which is clearly true. We have shaded the correct side of the line.
Answer:
x = 21
Step-by-step explanation:
1. Write the equation
(18 + 2x) ÷ 5 = 12
2. Multiply both sides by 5
18 + 2x = 60
3. Subtract 18 from both sides
2x = 42
4. Solve for x by dividing both sides by 2
x = 21
Answer:
Yes it will be appropriate to model the distribution of a sample mean with a normal model
Step-by-step explanation:
Given that the population is not normal, and the sample is sufficiently large, according to the Central Limit theorem, the distribution of the mean pf the sampling distribution will be approximately normal not withstanding the population from which the sample is obtained. Therefore, the mean, 
, and the standard deviation, 
, of the sample will be equal to the mean, μ, and standard deviation, σ, of the of the population
Therefore, it will be appropriate to model the distribution of a sample mean with a normal model
