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aksik [14]
3 years ago
14

-13=5(1+4x)-2x With steps

Mathematics
2 answers:
Allushta [10]3 years ago
8 0
-13=5+20x-2x
-13=5+18x
-18=18x
x= -1


patriot [66]3 years ago
7 0

How to get answer by Mimiwhatsup:

\mathrm{Switch\:sides}\\5\left(1+4x\right)-2x=-13\\\mathrm{Expand}\:5\left(1+4x\right):\quad 5+20x\\5+20x-2x=-13\\\mathrm{Add\:similar\:elements:}\:20x-2x=18x\\5+18x=-13\\\mathrm{Subtract\:}5\mathrm{\:from\:both\:sides}\\5+18x-5=-13-5\\Simplify\\18x=-18\\\mathrm{Divide\:both\:sides\:by\:}18\\\frac{18x}{18}=\frac{-18}{18}\\Answer: x=-1

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Paguei 74,00 por uma bolsa e uma sandália. A bolsa foi 23,00 mais barato q a sandália. Qual o preço da sandália?
alukav5142 [94]

Responda:

48,5

Explicação passo a passo:

Dado que:

Custo total de bolsa e sandália = 74,00

Deixei :

Custo da sandália = x

Custo da bolsa = x - 23

Preço da sandália:

x + (x - 23) = 74

x + x - 23 = 74

2x - 23 = 74

2x = 74 + 23

2x = 97

x = 97/2

x = 48,5

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Answer:

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Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

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3 years ago
A cake has 32 slices.
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