Ask question

Ask question

All categories

3 years ago

7

3.3 - 3.7n - 2n + 8 nnnnnnnnnnnnnnnnnn

2 answers:

qaws [65]3 years ago

5 0

enejjennnenenswatehd

arsen [322]3 years ago

3 0

Answer:

-5.7n + 11.3

Step-by-step explanation:

3.3 - 3.7n - 2n + 8 = -3.7n - 2n + 3.3 + 8 = -5.7n + 11.3

You might be interested in

3(y-2)=18<br> 3y-____=18<br> 3y=____

antoniya [11.8K]

Greetings!

<span>3(y-2)=18
Distribute the Parenthesis (3*y+3*-2).
3y-6=18
To Isolate the variable, we have to add/subtract any unlike terms.
3y=24

To solve for y.
</span> $3y=24$
<span>Divide both sides by 3.
</span> $(3y)/3=(24)/3$
<span>The Answer Is:
</span> $y=8$ <span>

Hope this helps.
-Benjamin
</span>

3 0

4 years ago

Read 2 more answers

You invest $1300 in an account at 2% per year simple interest. The equation

myrzilka [38]

Answer:

B. $1430

Step-by-step explanation:

Put 6 in the equation for n and do the arithmetic.

A(6) = 1300 +(6 -1)(0.02·1300)

= 1300(1 +5(.02)) = 1300(1.1)

A(6) = 1430 . . . . matches choice B

4 0

3 years ago

An someone help me please: (2x - 6) + 4(x - 3)

Alex777 [14]

Answer:

6x - 18

Step-by-step explanation:

thats it simplified! hope it helps!

6 0

3 years ago

Use the following prompt to answer the next 6 questions. Suppose we want to test the color distribution claim on the M&M’s w

vitfil [10]

Answer:

The claim on the M&M’s website is not true.

Step-by-step explanation:

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

<em>H</em>₀: The observed frequencies are same as the expected frequencies.

<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

$\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}$

Here,

$O_{i}$ = Observed frequencies

$E_{i}=N\times p_{i}$ = Expected frequency.

The chi-square test statistic value is, 14.433.

The degrees of freedom is:

df = <em>k</em> - 1 = 6 - 1 = 5

Compute the <em>p</em>-value as follows:

$p-value=P(\chi^{2}_{k-1} >14.433) =P(\chi^{2}_{5} >14.433) =0.013$

*Use a Chi-square table.

The significance level is, <em>α</em> = 0.05.

p-value = 0.013 < <em>α</em> = 0.05.

So, the null hypothesis will be rejected at 5% significance level.

Thus, concluding that the claim on the M&M’s website is not true.

6 0

3 years ago

Tony walks 6/8 mile every day. How far does tony walk in 5 days ?

Gennadij [26K]

I believe it would be 3.75 miles in 5 days.....

hope this helped!!!!

8 0

3 years ago