Answer: none
Step-by-step explanation:
(A)
(16÷32/10) ×2 + 0.2×(90)
Using bodmas principle ; solve bracket
(16×10/32)×2 + (2/10×90)
10+18 =28
(B)
{(16÷32/10) × (2+2/10)} ×90
Open brackets
{(16×10/32) × (22/10)} ×90
(5×11/5) ×90
11×90 = 990
(C)
16÷{(32/10×2) + (2/10×8)} +82
Open brackets, solve division first, dolled by addition
16÷(32/5 + 8/5) +82
16÷(40/5) +82
16÷8 +82
2+82= 84
(D)
[16÷(32/10 ×2) + 0.2× (90)]
16÷ (32/5) + 2/10 ×90
Solve division
16×5/32 + 18
5/2 + 18
L.c.m of denominator (2&1) =2
(5+36) / 2 = 41/2
=20.5
Answer:
can u please add the statements.
Step-by-step explanation:
Answer:
360 learners
Step-by-step explanation:
One way to solve these types of problems is that you could set up a proportion
learners/teachers = learners/teachers
Let x = no. of learners to enrol
30/1 = x/12
12(30) = x
x = 360 learners
Answer:
The claimed proportion (population) is
The sample proportion (p-hat) is
Step-by-step explanation:
In the population, we have the parameter p that is 11%.
Then, a sample of n=160 is taken and the proportion of the sample is 0.10.
The sample proportion p-hat can differ from the population's proportion. There is a sampling distribution that gives the possible values of the sample proportion taken out of this population and its associated probabilities.
If triangle ABC is congruent to triangle DEF, then EF = BC = 27
This is because BC and EF are the last two letters of ABC and DEF respectively. They match up and correspond, being congruent by CPCTC
-------------------------------------------------------------------------------------------------------------
Similarly, if triangle ABC is congruent to triangle DEF, then angle D = angle A = 49 degrees
The letter D and the letter A are the first letters of DEF and ABC respectively. So they match up and are congruent by CPCTC
CPCTC = Corresponding Parts of Congruent Triangles are Congruent
-------------------------------------------------------------------------------------------------------------
So in short, the answer is choice B) 27; 49
