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zhuklara [117]
3 years ago
10

11) Write the explicit formula

Mathematics
1 answer:
Illusion [34]3 years ago
6 0

Answer:

365

Step-by-step explanation:

tn = t1 + (n - 1)(d)

t(47) = -3 + (47 - 1)(8)

t(47) = -3 + (46)(8)

t(47) = -3 + 368

t(47) = 365

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Pleas help me (and pls don't ban my question this time Brainly)
Evgesh-ka [11]

Answer:

d = 7.2

Step-by-step explanation:

12 + d = -4.8

subtract 12 from both sides and you get

d = 7.2

5 0
3 years ago
Read 2 more answers
9(2.3n+6)+10.45>43.7
kifflom [539]

Answer:

Step-by-step explanation:

9(2.3n+6)+10.45>43.7

(9*2.3n)+(9*6)+10.45>43.7

20.7n+54+10.45>43.7

20.7n+64.45>43.7

7 0
3 years ago
There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit
DaniilM [7]

Answer: p = 0.9337

Step-by-step explanation: from the question, we have that

total number of pen (n)= 15

number of pen that has never been used=9

number of pen that has been used = 15 - 9 =6

number of pen choosing on monday = 3

total number of pen choosing on tuesday=3

note that the total number of pen is constant (15) since he returned the pen back .

probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

probability of picking a pen that has been used on tuesday = 6/15 = 2/5

probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5

on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .

we know that

probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.

to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution

p(x=r) = nCr * p^{r} * q^{n-r}

n= total number of pens=15

r = number of pen chosen on tuesday = 3

p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

by slotting in the parameters, we have that

p(x=3) = 15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}

p(x=3) = 455 * 0.4^{3} * 0.6^{12}

p(x=3) = 455 * 0.064 * 0.002176

p(x=3) = 0.0633

thus probability that 3 of the pens has been used on tuesday. = 0.0633

probability of ball that none of the 3 pen has never being used on tuesday  = 1 - 0.0633 = 0.9337

3 0
3 years ago
4+2(7)= <br><br> 62+8×3= <br><br> 3(2+5)−5(3)+8= <br><br> 24−6+12÷2×3=
Temka [501]

4+2(7)= 18

62+8x3= 86

3(2+5)-5(3)+8= 14

24-6+12÷2x3= 36

Hope this helps you! :D

3 0
4 years ago
What is the sum? (1/x+2)+(1/x+3)+(1/X^2+5+6)
PolarNik [594]
For this case we have the following expression:
 (1 / x + 2) + (1 / x + 3) + (1 / X ^ 2 + 5 + 6)
 Rewriting we have:
 (1 / x + 2) + (1 / x + 3) + (1 / ((x + 2) * (x + 3)))
 By doing common factor we have:
 (1 / ((x + 2) * (x + 3))) * (x + 3 + x + 2 + 1)
 Rewriting:
 (1 / ((x + 2) * (x + 3))) * (2x + 6)
 The sum is:
 ((2x + 6) / ((x + 2) * (x + 3)))
 Answer:
 
((2x + 6) / ((x + 2) * (x + 3)))
7 0
3 years ago
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