All categories

3 years ago

Calculate the mean of the electric bills for Family C. Round your answer to the nearest cent. $

Mathematics

1 answer:

Dmitry [639]3 years ago

8 0

Is there a paragraph or a chart?

You might be interested in

Check

asambeis [7]

Answer:

Step-by-step explanation:

An equation contains a sign equality (=) between two expressions.

And an expression is a simple algebraic statement.

By these definitions,

Equations Expressions

x = 2x + 5 21x - 13 + 11x

$\frac{1}{2}x=5+3$ -3.7x - 13

3x + 1 = -2.1 $\frac{1}{2}x+5$

7 0

3 years ago

What is 2.9 as a fraction

Pani-rosa [81]

Answer:

$\frac{29}{10}$

Step-by-step explanation:

8 0

3 years ago

When the denominator and numerator are equal, it means?

WITCHER [35]

The fraction is one whole aka one

5 0

3 years ago

M,H, and G have written more than a combined total of 22 articles. H has written 1/4 as many as M. G has written 3/2 as M. Write

Natalka [10]

Missing part of the question:

Write an inequality to determine the number of articles, M could have written for the school newspaper.

Answer:

The inequality: $11M > 88$

The solution: $M > 8$

Step-by-step explanation:

Given

From the question, we have the following parameters:

$M + H + G > 22$

$H = \frac{1}{4}M$

$G = \frac{3}{2}M$

Required

Determine the inequality to solve for M

Substitute the values for H and G in the inequality:

$M + H + G > 22$

$M + \frac{1}{4}M + \frac{3}{2}M > 22$

Multiply through by 4

$4(M + \frac{1}{4}M + \frac{3}{2}M) > 22*4$

$4M + M + 6M > 88$

$11M > 88$

Divide both sides by 11

$M > 8$

4 0

3 years ago

A mixture of compounds X and Y in a 0.100-cm cell had an absorbance of 0.215 at 272 nm and 0.191 at 327 nm. Find [X] and [Y] in

Oksi-84 [34.3K]

Answer: The concentration of X is $7.13582\times 10^{-6}$ and the concentration of Y is $2.53159\times 10^{-5}$ .

Step-by-step explanation:

Since we have given that

At 272 nm, absorbance = 0.215

At 327 nm, absorbance = 0.191

As we have given that

Compound X Compound Y

272 16400 3870

327 3990 6420

So, our equations becomes

$16400C_1+3870c_2=0.215\\\\3990C_1+6420C_2=0.191$

By solving these two equations, we get that

$C_1=7.13582\times 10^{-6}\\\\C_2=2.53159\times 10^{-5}$

Hence, the concentration of X is $7.13582\times 10^{-6}$ and the concentration of Y is $2.53159\times 10^{-5}$ .

5 0

4 years ago