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3 years ago

Use the Distributive Property and simplify the expression. 6+4(3x – 2)+5

1 answer:

8 0

First, we need to use the distributive property to simplify the expression. So, that means we have to multiply the 4 by all of the integers and variables inside the parentheses.

If done correctly, your expression should be 6+12x-8+5

Next, we should combine all of the constant terms (just numbers without variables), also known as combining like terms.

If done correctly, your expression should be 12x+3.

This matches answer B. 12x+3

You might be interested in

Find the angle of elevation if the building is 653 feet tall and the observer is 150 feet from the building. Round to the neares

natita [175]

Answer:

Angle of elevation = 77.1°

Step-by-step explanation:

The formula to solve for angle if elevation is given below as:

tan θ = Opposite/ Adjacent

Where

Opposite = Height of the building = 653 ft

Adjacent = Distance from the building = 150 ft

θ = Angle of Elevation = ?

Hence:

tan θ = 653ft/150ft

θ = arc tan (653ft/150ft)

= 77.063069956°

Approximately to the nearest tenth = 77.1°

6 0

3 years ago

(a) the number 561 factors as 3 · 11 · 17. first use fermat's little theorem to prove that a561 ≡ a (mod 3), a561 ≡ a (mod 11),

Vitek1552 [10]

LFT says that for any prime modulus $p$ and any integer $n$ , we have

$n^p\equiv n\pmod p$

From this we immediately know that

$a^{561}\equiv a^{3\times11\times17}\equiv\begin{cases}(a^{11\times17})^3\pmod3\\(a^{3\times17})^{11}\pmod{11}\\(a^{3\times11})^{17}\pmod{17}\end{cases}\equiv\begin{cases}a^{11\times17}\pmod3\\a^{3\times17}\pmod{11}\\a^{3\times11}\pmod{17}\end{cases}$

Now we apply the Euclidean algorithm. Outlining one step at a time, we have in the first case $11\times17=187=62\times3+1$ , so

$a^{11\times17}\equiv a^{62\times3+1}\equiv (a^{62})^3\times a\stackrel{\mathrm{LFT}}\equiv a^{62}\times a\equiv a^{63}\pmod3$

Next, $63=21\times3$ , so

$a^{63}\equiv a^{21\times3}=(a^{21})^3\stackrel{\mathrm{LFT}}\equiv a^{21}\pmod3$

Next, $21=7\times3$ , so

$a^{21}\equiv a^{7\times3}\equiv(a^7)^3\stackrel{\mathrm{LFT}}\equiv a^7\pmod3$

Finally, $7=2\times3+1$ , so

$a^7\equiv a^{2\times3+1}\equiv (a^2)^3\times a\stackrel{\mathrm{LFT}}\equiv a^2\times a\equiv a^3\stackrel{\mathrm{LFT}}\equiv a\pmod3$

We do the same thing for the remaining two cases:

$3\times17=51=4\times11+7\implies a^{51}\equiv a^{4+7}\equiv a\pmod{11}$

$3\times11=33=1\times17+16\implies a^{33}\equiv a^{1+16}\equiv a\pmod{17}$

Now recall the Chinese remainder theorem, which says if $x\equiv a\pmod n$ and $x\equiv b\pmod m$ , with $m,n$ relatively prime, then $x\equiv b{m_n}^{-1}m+a{n_m}^{-1}n\pmod{mn}$ , where ${m_n}^{-1}$ denotes $m^{-1}\pmod n$ .

For this problem, the CRT is saying that, since $a^{561}\equiv a\pmod3$ and $a^{561}\equiv a\pmod{11}$ , it follows that

$a^{561}\equiv a\times{11_3}^{-1}\times11+a\times{3_{11}}^{-1}\times3\pmod{3\times11}$
$\implies a^{561}\equiv a\times2\times11+a\times4\times3\pmod{33}$
$\implies a^{561}\equiv 34a\equiv a\pmod{33}$

And since $a^{561}\equiv a\pmod{17}$ , we also have

$a^{561}\equiv a\times{17_{33}}^{-1}\times17+a\times{33_{17}}^{-1}\times33\pmod{17\times33}$
$\implies a^{561}\equiv a\times2\times17+a\times16\times33\pmod{561}$
$\implies a^{561}\equiv562a\equiv a\pmod{561}$

6 0

4 years ago

A researcher classifies firefighters according to whether their gloves fit well or poorly and by gender. They want to know if th

snow_tiger [21]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the proportion of "X: gloves fitness, categorized: Fit poorly and Fit well" is the same for two populations of interest, "male firefighters" and "female firefighters"

To do this you have to conduct a Chi-Square test of Homogeneity.

In the null hypothesis you have to state that the proportion of the categories of the variable are the same for all the populations of interest.

M: the firefighter is male

F: the firefighter is female

Y: represents the category that the gloves "fit poorly"

W: represents the category that the gloves "fit well"

The null hypothesis will be:

H₀: P(Y|M)=P(Y|F)=P(Y)

P(W|M)=P(W|F)=P(W)

H₁: At least one of the statements in the null hypothesis is false.

α: 0.01

To calculate the statistic under the null hypothesis you have to calculate the expected frequencies first:

$E_{ij}= O_{.j}*\frac{O_{i.}}{n}$

O.j= total of the j-column

Oi.= total of the i-row

n= total of observations

$E_{11}= 547*\frac{152}{586} = 141.88$

$E_{12}=39*\frac{152}{586}= 10.12$

$E_{21}= 547*\frac{434}{586} = 405.12$

$E_{22}= 39*\frac{434}{586} = 28.88$

$X^2= sum \frac{(O_{ij}-E_{ij})^2}{E_{ij}} ~~~X^2_{(r-1)(c-1)}$

r= number of rows (in this case 2)

c=number of columns (in this case 2)

$X^2_{H_0}= \frac{(132-141.88)^2}{141.88} +\frac{(20-10.12)^2}{10.12} +\frac{(415-405.12)^2}{405.12} +\frac{(19-28.88)^2}{28.88} = 13.95$

Using the critical value approach, you have to remember that this test is always one-tailed to the right, meaning that you'll have only one critical value from which the rejection region is defined:

$X^2_{(r-1)(c-1);1-\alpha }= X^2_{1;0.99}= 6.635$

The decision rule is then:

If $X^2_{H_0}$ ≥ 6.635, reject the null hypothesis.

If $X^2_{H_0}$ < 6.635, do not reject the null hypothesis.

The calculated value is greater than the critical value, the decision is to reject the null hypothesis.

So at a 1% level you can conclude that this test is significant. This means that the proportions of gloves fitness, categorized in "Fit poorly" and "Fit well" are different for the male and female firefighters populations.

I hope this helps!