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Katen [24]
3 years ago
12

PLEASE HELP ME !! ANSWER EACH QUESTION (2)

Mathematics
1 answer:
ss7ja [257]3 years ago
8 0
10)
If m<1 > m<2 then BC > CD

Answer
C.   BC > CD

----------------
11)
If BC >= EF then <BAC >= <EDF

Answer
C.  <BAC >= <EDF
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Any and all answers will get a 5 star rating
Afina-wow [57]
Sorry about that i didnt read the question properly

3 0
3 years ago
Read 2 more answers
A population of 55 foxes in a wildlife preserve doubles in size every 14 years. The function y equals 55 times 2 Superscript x ​
Pavlova-9 [17]
Since <em>x</em> is the number of 14-year periods, the first thing we do is divide 28 years by 14. 28/14=2.  Now we use that in our function, y=55*2^x.

y=55*2^2=55*4=220
There will be 220 foxes after 28 years.
6 0
3 years ago
Shenelle has 100100100 meters of fencing to build a rectangular garden. the garden's area (in square meters) as a function of th
cluponka [151]
Area as a function of width, w is:
a(w)=(w-25)^2+625
for maximum area, the width will be:
a'(w)=2(w-25)+0=0
solving for w we egt
2w-50=0
2w=50
w=25 m
given that the perimeter is 100, the length will be:
100=2(L+W)
solving for L we get:
L=50-W
but W=25m
hence
L=50-25=25 m
thus the maximum area will be:
A=L*W=25*25=625m^2

8 0
3 years ago
Read 2 more answers
Evaluate 5j +12 foreach listed<br> value ofn.<br> j= 4<br> j = 6
Bezzdna [24]

Answer: 32 and 42

Step-by-step explanation:

5j +12

j= 4

j = 6

5(4) + 12

5 times 4 = 20

20 + 12 = 32

5(6) + 12

5 times 6 = 30

30 + 12 = 42

6 0
3 years ago
Read 2 more answers
The monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be normally distribu
omeli [17]

Answer:  0.9996

Step-by-step explanation:

Given : The monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be normally distributed with  a mean \mu=$\ 87.00

Standard deviation : \sigma=$\ 36.00

Sample size : n=100

Let X be the random variable that represents the electricity utility bill for a randomly selected month .

z-score : z=\dfrac{X-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For X = $75.00

z=\dfrac{75.00-87.00}{\dfrac{36}{\sqrt{100}}}\approx-3.33

Now, the probability that the average bill for those sampled will exceed $75.00 will be :-

P(X>75)=P(z>-3.33)=1-P(z\leq-3.33)\\\\=1-  0.0004342=0.9995658\approx0.9996

Hence, the probability that the average bill for those sampled will exceed $75.00 =0.9996

3 0
4 years ago
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