Jaime wrote 4.4_0.33=1.1 is his answer reasonable?why're why not?

1 answer:

7 0

No you have to make sure to line up the decimals when subtracting them. Do not just line of the numbers, this wills get you the wrong answers. Hope this helps. If you need anything else just call for a helping hand.

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PLZZ HELP!! WILL MARK BRAINLIEST PLZZ!!!

grandymaker [24]

Answer:

see below

Step-by-step explanation: 7 1 16 33

y = x² translated t the point (3, 2) y = 0 when x = 0

y = (x - 3)² moves the function three units to the right y = 0 when x = -3

y = (x-3)² + 2 moves the function up 2 units y = 2 when x = -3

y = (x-3)² + 2

y = x² - 6x + 9 + 2

y = x² - 6x +11

graph the equations x², (x-3)² + 2, and x² - 6x + 11 to verify (I did)

7 0

3 years ago

Write an equation (in factor form) that has solutions of x=2 or x=3

Art [367]

This means that when you write your equation you will have in x somewhere in it the X will be the number 2 or the number 3. hope thia helps good luck!

5 0

4 years ago

Read 2 more answers

Construct a 95% confidence interval for the population mean, mu. Assume the population has a normal distribution. A random sam

aniked [119]

Given Information:

Number of lithium batteries = n = 16

Mean life of lithium batteries = μ = 645 hours

Standard deviation of lithium batteries = σ = 31 hours

Confidence level = 95%

Required Information:

Confidence Interval = ?

Answer:

$CI = 628.5 \: to \: 661.5 \: hours$

Step-by-step explanation:

The confidence interval is given by

$CI = \mu \pm t_{\alpha/2} (\frac{\sigma}{\sqrt{n}})$

Where μ is the mean life of lithium batteries, σ is the standard deviation, n is number of lithium batteries selected, and t is the critical value from the t-table with significance level of

tα/2 = (1 - 0.95) = 0.05/2 = 0.025

and the degree of freedom is

DoF = n - 1 = 16 - 1 = 15

The critical value (tα/2) at 15 DoF is equal to 2.131 (from the t-table)

$CI = 645 \pm 2.131(\frac{31}{\sqrt{16}})$