Solve for x. Show the equation you used and your work. <br><br> Question 2 options:

uysha [10]

$Use:\\logb^c=c\cdot log b\\\\log(a\cdot b)=loga+logb\\\\==================================$

$5)\\log270=log(27\cdot10)=log27+log10=log3^3+1=3log3+1=\boxed{3u+1}\\\\or\\=log27+log10=log(3\cdot9)+1=log3+log9+1=\boxed{u+s+1}$

$6)\\log99=log(9\cdot11)=log9+log11=log3^2+t=2log3+t=\boxed{2u+1}\\or\\=log9+log11=\boxed{s+t}$

$7)\\log121=log11^2=2log11=\boxed{2t}$

$8)\\log363=log(121\cdot3)=log121+log3=log11^2+u=2log11+u=\boxed{2t+u}$

7 0

3 years ago

Read 2 more answers

In the algebra expression 7a+17b-10c,7, 17, and -10 are

AVprozaik [17]

These numbers are categorized as numerical coefficients. A number that is next to a specific variable in an algebraic or polynomial expression and is essentially being multiplied to that variable.

6 0

3 years ago

1. Kayla wants to find the width, AB, of a river. She walks along the edge of the river 100 ft and marks point C. Then she walks

Zolol [24]

You are given Kayla's walking distance along the edge of the river 100 ft and marks point C, then she walks 100 ft further and marks point D and she turns 90° and walks until her location, point A, and point C are collinear. She marks point E at this location. ABC and EDC are congruent. thats just what i think it is tho

6 0

3 years ago

Suppose that a jewelry store tracked the amount of emeralds they sold each week to more accurately estimate how many emeralds to

RUDIKE [14]

Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

5 0

3 years ago

Spiral Review

Svetllana [295]

Answer:

Your answer is C

Step-by-step explanation:

0.5 = Five tenths

and

0.05 = five hundredths

This is because 0.5 is like 50 cents and 0.05 is like 5 cents

The rest are wrong because A: 0.05 simply can't be bigger than 0.5

B is wrong because they aren't equal to each other

and D is wrong because 0.05 + 0.5 = 0.55

(\) QueTooOfficial (/)

3 0

3 years ago

Read 2 more answers

Solve for x. Show the equation you used and your work. Question 2 options: