1. A book has x pages. How many pages are in the book if Maria read 45 pages of a book on

ASAAPPPPPP!!!!! Complete the tables of values

velikii [3]

Answer:

see explanation

Step-by-step explanation:

Using the rules of exponents

$a^{-m}$ ⇔ $\frac{1}{a^{m} }$ and $a^{0}$ = 1

Substitute the values of x into $4^{-x}$

x = 0 → $4^{0}$ = 1 ⇒ a = 1

x = 2 → $4^{-2}$ = $\frac{1}{4^{2} }$ = $\frac{1}{16}$ , hence

b = $\frac{1}{16}$

x = 4 → $4^{-4}$ = $\frac{1}{4^{4} }$ = $\frac{1}{256}$ , hence

c = $\frac{1}{256}$

6 0

3 years ago

For each of the -3.3 trials, and in the same sequence in which you entered the freezing points, enter your calculated values for

dedylja [7]

Answer:

The morality of the solution is calculated as 0.859 m. We are required to determine the freezing point depression constant of pure water. The freezing point depression of the solution is given as

$T_{f} = T_{p} -T_{s} = K_{f}$ * (morality of solution)

$T_{p}$ and $T_{s}$ are the freezing points of the pure solvent (water, 0°C) and $K_{f}$ = freezing point depression constant of water. Therefore,

$T_{f} = K_{f}$ *(0.859 m)

=====> (0°C) – (3.00°C) = $K_{f} *(0.859 m)$

=====> -3.00°C = $K_{f} *(0.859 m)$

Ignore the negative sign (since $K_{f}$ is positive) and get

$K_{f}$ = (3.00°C) / (0.859 m) = 3.492°C/m

The freezing point depression constant of the solvent is 3.5°C kg/mole

3.5 K.kg/mole (temperature differences are the same in Celsius and Kelvin scales).

3 0

3 years ago

Helpppppp meeeee pleaseeeee

Novay_Z [31]

Answer:

B) -2 degrees

Step-by-step explanation:

If the temperature went down 10 degrees the equation you need to use is 8-10=x.

Once simplified/solved you will get -2

7 0

3 years ago

Read 2 more answers

If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2

nikklg [1K]

Answer:

n=97

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=0.01)$