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Alexandra [31]
4 years ago
7

2. In a certain year, solar and wind energy generation totaled 1037 Trillion Btu’s. Wind generated 56 Trillion Btu's more than e

ight times that generated by solar energy. How much was generated by each source?
Mathematics
1 answer:
GaryK [48]4 years ago
5 0
The sum of solar energy can assumed as 'x', so it is easy to conclude that the sum of wind energy is '8x+56'. From the above assumption, we can easily get that 'x+8x+56=1037', to solve thi<span>s equation and get tha</span>
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Please help me !!!!!!
cestrela7 [59]

Given:

The set of pair and graphs.

To find:

The domain and range.

Solution:

We know that,

Domain is the set of x-values or input values.

Range is the set of y-values or output values.

(a)

The given set of ordered pairs is

{(-3,3),(5,5),(-3,2),(5,3)}

Here, the x-coordinates are -3, 5, -3, 5.

A set contains distinct values.

Therefore, the domain is {-3,5}.

(b)

The graph is given.

From the given graph the set of ordered pairs is

{(-2,1),(-1,0.5),(-1,3),(0,0),(0,2),(1,0.5),(1,3)(2,1)}

Here, the y-values are 1, 0.5, 3, 0, 2, 0.5, 3, 1.

Therefore, the range is {0, 0.5, 1, 2, 3}.

(c)

The graph is given.

From the given graph the set of ordered pairs is

{(-2,3),(-1,3),(0,1),(2,4)}

Here, the x-values are -2, -1,0, 2.

Therefore, the domain is {-2,-1,0,2}.

4 0
3 years ago
8j + 3-j=17<br> solve for J
givi [52]
First you have to combine like terms which you would combine 8j and j which would be 7j+3=17 then subtract 3 from 17 giving you 14. And after that your left with 7j=14 then divide 7 on both sides giving you 2 as your answer.
8 0
3 years ago
Read 2 more answers
Svetlana's hair is 4 centimeters long. Her hair grows 1.5, point, 5 centimeters per month. Svetlana wants her hair to grow so th
Alekssandra [29.7K]

4 < 5

I think that is the equality.

5 0
3 years ago
Read 2 more answers
Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a
kipiarov [429]

Answer:

a) P(k≤11) = 0.021

b) P(k>23) = 0.213

c) P(11≤k≤23) = 0.777

P(11<k<23) = 0.699

d) P(15<k<25)=0.687

Step-by-step explanation:

a) What is the probability that the number of drivers will be at most 11?

We have to calculate P(k≤11)

P(k\leq11)=\sum_0^{11} P(k

P(k=0) = 20^0e^{-20}/0!=1 \cdot 0.00000000206/1=0\\\\P(k=1) = 20^1e^{-20}/1!=20 \cdot 0.00000000206/1=0\\\\P(k=2) = 20^2e^{-20}/2!=400 \cdot 0.00000000206/2=0\\\\P(k=3) = 20^3e^{-20}/3!=8000 \cdot 0.00000000206/6=0\\\\P(k=4) = 20^4e^{-20}/4!=160000 \cdot 0.00000000206/24=0\\\\P(k=5) = 20^5e^{-20}/5!=3200000 \cdot 0.00000000206/120=0\\\\P(k=6) = 20^6e^{-20}/6!=64000000 \cdot 0.00000000206/720=0\\\\P(k=7) = 20^7e^{-20}/7!=1280000000 \cdot 0.00000000206/5040=0.001\\\\

P(k=8) = 20^8e^{-20}/8!=25600000000 \cdot 0.00000000206/40320=0.001\\\\P(k=9) = 20^9e^{-20}/9!=512000000000 \cdot 0.00000000206/362880=0.003\\\\P(k=10) = 20^{10}e^{-20}/10!=10240000000000 \cdot 0.00000000206/3628800=0.006\\\\P(k=11) = 20^{11}e^{-20}/11!=204800000000000 \cdot 0.00000000206/39916800=0.011\\\\

P(k\leq11)=\sum_0^{11} P(k

b) What is the probability that the number of drivers will exceed 23?

We can write this as:

P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))

P(k=12) = 20^{12}e^{-20}/12!=8442485.238/479001600=0.018\\\\P(k=13) = 20^{13}e^{-20}/13!=168849704.75/6227020800=0.027\\\\P(k=14) = 20^{14}e^{-20}/14!=3376994095.003/87178291200=0.039\\\\P(k=15) = 20^{15}e^{-20}/15!=67539881900.067/1307674368000=0.052\\\\P(k=16) = 20^{16}e^{-20}/16!=1350797638001.33/20922789888000=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=27015952760026.7/355687428096000=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=540319055200533/6402373705728000=0.084\\\\

P(k=19) = 20^{19}e^{-20}/19!=10806381104010700/121645100408832000=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=216127622080213000/2432902008176640000=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=4322552441604270000/51090942171709400000=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=86451048832085300000/1.12400072777761E+21=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=1.72902097664171E+21/2.5852016738885E+22=0.067\\\\

P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))\\\\P(k>23)=1-(0.021+0.766)=1-0.787=0.213

c) What is the probability that the number of drivers will be between 11 and 23, inclusive? What is the probability that the number of drivers will be strictly between 11 and 23?

Between 11 and 23 inclusive:

P(11\leq k\leq23)=P(x\leq23)-P(k\leq11)+P(k=11)\\\\P(11\leq k\leq23)=0.787-0.021+ 0.011=0.777

Between 11 and 23 exclusive:

P(11< k

d) What is the probability that the number of drivers will be within 2 standard deviations of the mean value?

The standard deviation is

\mu=\lambda =20\\\\\sigma=\sqrt{\lambda}=\sqrt{20}= 4.47

Then, we have to calculate the probability of between 15 and 25 drivers approximately.

P(15

P(k=16) = 20^{16}e^{-20}/16!=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=0.084\\\\P(k=19) = 20^{19}e^{-20}/19!=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=0.067\\\\P(k=24) = 20^{24}e^{-20}/24!=0.056\\\\

3 0
3 years ago
7. Find the measure of &lt;1. Justify your answer
ANTONII [103]

Answer:

3rd option

Step-by-step explanation:

∠ 1 and 75° are corresponding angles and are congruent , so

∠ 1 = 75°

8 0
3 years ago
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