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kupik [55]
2 years ago
12

If k= t-5k and k= 4, what is the value of t? A) -28 B) -12 C) 12 D) 28

Mathematics
2 answers:
Gre4nikov [31]2 years ago
8 0

Answer:

24

Step-by-step explanation:

k = t - 5k

→ Substitute in k

4 = t - 20

→ Add 20 to both sides

24 = t

alexgriva [62]2 years ago
7 0
The answer would be 28 i think
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Answer:

\large \boxed{\text{A) } \dfrac{6}{x} = \dfrac{9}{\$7.74}}

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According to the alternendo property,

\dfrac{9}{6} = \dfrac{\$7.74}{x} \Rightarrow \dfrac{9}{\$7.74} = \dfrac{6}{x}\\\\\text{The correct answer is $\large \boxed{\mathbf{ \dfrac{6}{x} = \dfrac{9}{\$7.74}}}$}

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g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
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Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0
3 years ago
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