Mathematics, 21.05.2020 00:08, phancharamachasm

How many solutions does the following equation have?
4(- 30) = 4y + 12
Choose 1
No solutions
B
Exactly one solution
Infinitely many solutions

Answers: 2

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Answers

answered: Guest

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺

answered: Guest

$2^{2x+11}=3^{x-31}$

take the logarithm of both sides (the base of the log doesn't matter):

$\ln2^{2x+11}=\ln3^{x-31}$

apply the exponent property - this says that $\ln a^b=b\ln a$ :

$(2x+11)\ln2=(x-31)\ln3$

distribute the log terms on both sides:

$(2\ln2)x+11\ln2=(\ln3)x-31\ln3$

group like terms together:

$(2\ln2-\ln3)x=-11\ln2-31\ln3$

divide through both sides by the coefficient on $x$ :

$x=-\dfrac{11\ln2+31\ln3}{2\ln2-\ln3}$

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like:

$x=-\dfrac{\ln2^{11}+\ln3^{31}}{\ln2^2-\ln3}$

$x=-\dfrac{\ln2^{11}3^{31}}{\ln\frac43}$

$x=-\log_{4/3}2^{11}3^{31}$

answered: Guest

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.

answered: Guest

if i did my math right then the answer should but 108