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3 years ago

Rearrange to make subject x 4(x-3)/a=y

Mathematics

1 answer:

VikaD [51]3 years ago

7 0

Answer:

answer is x = ay + 12/4

Step-by-step explanation:

first we cross multiply on both sides which will give us this

4( x -3)=ay

we then open the brackets by multiplying 4 with x and -3 which gives

4x - 12= ay

then take 12 to the other side

4x - 12+12= ay + 12

4x = ay + 12

divide 4 on both sides

4x /4= ay + 12/ 4

therefore x = ay + 12/ 4

clear steps in picture

hope that can help

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2 years ago

The probability that a grader will make a marking error on any particular question of a multiple-choice exam is 0.15. If there a

sashaice [31]

Answer:

$P(X=0)=(10C10)(0.15)^{0} (1-0.15)^{10-0}=0.1969$

$P(X \geq 1)= 1-P(X$

$P(X=0)=(nCn)(p)^{0} (1-p)^{n-0}=(1-p)^n$

$P(X \geq 1)= 1-P(X$

Step-by-step explanation:

Previous concepts

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The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.15)$

what is the probability that no errors are made?

For this case means that all the questions were correct so we want this probability:

$P(X=0)=(10C10)(0.15)^{0} (1-0.15)^{10-0}=0.1969$

what is the probability that at least one error made?

For this case we want this probability:

$P(X \geq 1)$

And we can use the complement rule:

$P(X \geq 1)= 1-P(X$

If there are n questions and the probability of a marking error is p rather than 0.15, give expressions for the probabilities of no errors and at least one error

$P(X=0)=(nCn)(p)^{0} (1-p)^{n-0}=(1-p)^n$

$P(X \geq 1)= 1-P(X$

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3 years ago

6% of what number is 24

Flura [38]

Answer:

4 i think, sorry if incorrect

Step-by-step explanation:

8 0

2 years ago

Can somebody help me What is 4(x-1)=6

avanturin [10]

4(x-1)=6 you start by dividing the whole equation by 4, so
x-1=6/4 simplify, x-1=3/2 then you add 1 to the whole equation
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7 0

3 years ago

A town has 5000 people in year t = 0. Calculate how long it takes for the population P to double once, twice, and three times, a

UNO [17]

Answer:

Step-by-step explanation:

At the time t = 0, population of the town = 5000

Rate of population increase = 500 per year

Therefore, the equation that will represent the population will be

$P_{t}=P_{0}+500t$

Where $P_{t}$ = Population after t years

$P_{0}$ = Initial population

t = Time in years

a). For double once the population will be 500×2 = 10000

By plugging in the values in the equation,

10000 = 5000 + 500t

500t = 10000 - 5000

500t = 5000

t = $\frac{5000}{500}$

t = 10 years

For Double twice,

Population will be = 10000×2 = 20000

Now we plug in the values in the equation again

20000 = 5000 + 500t

500t = 20000 - 5000

500t = 15000

t = $\frac{15000}{500}$

t = 30 years

For double thrice,

Population of the town = 20000×2 = 40000

Now we plug in the values in the equation,

40000 = 5000 + 500t

500t = 40000 - 5000

500t = 35000

t = $\frac{35000}{500}$

t = 70 years

b). If the population growth is 5%.

Then the growth will be exponential represented by

$T_{n}=T_{0}(1+\frac{r}{100})^{t}$

$T_{n}$ = Population after t years

$T_{0}$ = Initial population

t = time in years

For double once,

Population after t years = 10000

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$(1.05)^{t}=\frac{10000}{5000}$

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Take log on both the sides

$log(1.05)^{t}=log2$

tlog(1.05) = log2

t = $\frac{log2}{log1.05}$

t = 14.20 years

For double twice,

Population after t years = 20000

$20000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{20000}{5000}$

$(1.05)^{t}=4$

Take log on both the sides

$log(1.05)^{t}=log4$

tlog(1.05) = log4

t = $\frac{log4}{log1.05}$

t = 28.413 years

For double thrice

Population after t years = 40000

$40000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{40000}{5000}$

$(1.05)^{t}=8$

Take log on both the sides

$log(1.05)^{t}=log8$

tlog(1.05) = log8

t = $\frac{log8}{log1.05}$

t = 42.620 years

4 0

2 years ago