4.23×10 to the third power in standard notation/4.23×10 to the third power equals 4.23 times blank equals blank

A bowling alley charges $8 for shoes and $3 per game Claire brought $20 what inequality describes this

SSSSS [86.1K]

Answer:

20 <u>></u> 8+3g

Step-by-step explanation:

7 0

2 years ago

. Express each exponent in expanded form: StartFraction 6 times 6 times 6 times 6 times 6 times 6 times 6 Over 6 times 6 times 6

vovangra [49]

Answer:

The exponent of the simplified power is 3

Step-by-step explanation:

Given:

6 × 6 × 6 × 6 × 6 × 6 × 6 / 6 × 6 × 6 × 6

Dividing common factors ( cancel out 4 6's in the Numerator and Denominator

= 6 × 6 × 6 / 1

= 6^3 /1

= 6^3

The exponent of the simplified power is 3

3 0

2 years ago

How to solve 2.25t+5=13.5t+14

anyanavicka [17]

2.25t+5= 13.5t+14
-2.25t -2.25t
5= 11.25t+14
-14 -14
-9= 11.25t
-9/11.25=t
t= -0.8

3 0

3 years ago

Suppose the number of inches of rainfall each year in a city is normally distributed. For a random sample of years, the confiden

quester [9]

Answer:

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

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

Or equivalently:

$\bar X \pm ME$

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

$ME= \frac{Upper -Lower}{2}= \frac{7.7-3.9}{2}= 1.9$

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 for the sample

$\mu$ population mean (variable of interest)

Solution to the problem

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

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

Or equivalently:

$\bar X \pm ME$

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

$ME= \frac{Upper -Lower}{2}= \frac{7.7-3.9}{2}= 1.9$

4 0

2 years ago

Anna opens two savings accounts at the beginning of the year. She deposits $5,000 in an account that earns 2% interest each year

xenn [34]

As we know she wont be depositing or withdrawing that means that the total money remaining in the bank would be static and would just add up the interest their is.

SO 2 per cent of 5000 is 100 $. So she will earn a 100 $ from her first account.

And 2.5 % from 10000 is around 250 $ so she earn a total of 250 $ from her second account.

Now adding up the complete money that she will earn at the end of the year would be around 350 $.

5 0

2 years ago