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1 year ago

311, 305,299,...Find the 32nd term.

Mathematics

1 answer:

adell [148]1 year ago

4 0

Answer:

503

Step-by-step explanation:

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A friend needs at least $125 to go on the class trip. He has saved $45. He makes

sammy [17]

Answer:

Sorry, I don't know exactly what an inequality is but he needs to mow 4 lawns.

Step-by-step explanation:

$125-$45=80

$20x?=$80

?=4

8 0

2 years ago

Read 2 more answers

Ryan is trying a low-carbohydrate diet. He would like to keep the amount of carbs consumed in grams between the levels shown in

nydimaria [60]

Answer:

$50$

Ryan would like to eat more than 50 carbs per day, but no more than 150 carbs per day.

So, Ryan's total carb intake must be between 50 and 150 carbs.

Step-by-step explanation:

So he wants to keep his consumption of carbs between the inequalities:

$110$

So, let's solve both inequalities.

$110$

Subtract 10 from both sides:

$100$

Divide both sides by 2:

$50$

$2x+10$

Subtract 10 from both sides:

$2x$

Divide both sides by 2:

$x$

So, our inequality is now:

$110$

Since we solved the equations:

$50$

Written as a compound inequality, this is:

$50$

In other words, Ryan would like to eat more than 50 carbs per day, but no more than 150 carbs per day.

So, Ryan's total carb intake must be between 50 and 150 carbs.

4 0

2 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

2 years ago

In ΔNOP, \overline{NP} NP is extended through point P to point Q, \text{m}\angle OPQ = (9x-19)^{\circ}m∠OPQ=(9x−19) ∘ , \text{m}

kow [346]

Answer:

To find the measure of angle OPQ, we have two equations...

y+5x-17=180 because angles OPN and OPQ are supplementary...

AND

x+17+2x-4+y=180 because the total degrees of a triangle must add up to 180.

The equations...

5x+y=197

3x+y=167

solce the system of equations...

x=15

y=122

Since we wanted to solve for y, 122 is the answer.

The measure of angle OPQ is 122.

3 0

1 year ago

Tell us how much you're losing in this example:

Sergio [31]

Answer:

My gain in that year is - $ 176.5

Step-by-step explanation:

I have $ 1,500 in my investment account. Rate of interest for a year in my investment account = 8.5 %.

So, the amount of money I am making on my investment in a year,

= $ $\frac {1500 \times 8.5}{100}$

= $ 127 .5

I owe $ 1600 on a credit card . Rate of interest a year on this credit card debt, = 19 % .

So, the money I am paying in interest in a year on my card,

= $ $\frac {1600 \times 19}{100}$

= $ 304

Since, neither I withdraw money from my investment account except the interest , nor I pay my whole credit card debt except the interest, so, my gain in that year,

= $ (127.5 - 304)

= - $ 176.5

So, my loss in that year is $ 176.5

7 0

2 years ago