Evaluate. 43 – 4:25 ο ο ο ο

Mathematics

1 answer:

meriva3 years ago

4 0

This is an improper. Perhaps you can fix it, so that I can assist you with it? I apologise.

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likoan [24]

X + 7, ie the number set on Monday + seven more

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3 years ago

Find the numerical value of the log expression. \log a=7\hspace{15px} \log b=10\hspace{15px} \log c=3 loga=7logb=10logc=3 \log \

bija089 [108]

Answer:

Step-by-step explanation:

$\log a=7\hspace{15px} \log b=10\hspace{15px} \log c=3\\\\ \log \dfrac{a^8c^2}{b^3}=8\log{a}+2\log{c}-3\log{b}=(8)(7)+(2)(3)-(3)(10)\\\\=56+6-30=\boxed{32}$

The applicable rules of logarithms are ...

log(ab/c) = log(a) +log(b) -log(c)

log(a^b) = b·log(a)

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3 years ago

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.