Fast answeer plssss thank you thank you

kolezko [41]

Step-by-step explanation:

$\tt{9 {x}^{2} - 12x + 4}$

Here, the second order of polynomial ax² + bx + c is factorized and expressed as the product of two linear factors. First, we have to find the two numbers that adds to 12 and if we multiply those numbers , we get 36 ( The two numbers should be 6 and 6 ).

⟶ $\tt{9 {x}^{2} - (6 + 6)x + 4}$

⟶ $\tt{9 {x}^{2} - 6x - 6x + 4}$ { Distribute x through the parentheses )

⟶ $\tt{ \underbrace{9 {x}^{2} - 6x}} \: - \underbrace{6x + 4}$

⟶ $\tt{3x(3x - 2) - 2(3x - 2)}$

⟶ $\tt{(3x - 2)(3x - 2)}$

⟶ $\tt{ {(3x - 2)}^{2} }$

$\pink{ \boxed{ \boxed{ \tt{ Our \: final \: answer : \boxed{ \underline{ \tt{{(3x - 2)}^{2} }}}}}}}$

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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4 0

3 years ago

Regina went out for breakfast. The bill came to $7.12. If Regina has a discount coupon for 20%, how much will be taken off the b

PolarNik [594]

I believe $1.42 would be taken off of the bill:))

3 0

3 years ago

Read 2 more answers

At a hardware store they can cut a rope off of a big roll so you can buy any length you like. the cost of 6 feet of rope is 7.50

konstantin123 [22]

Answer:

first you'd need to divide 7.50 by 6 which will get you (7.50÷6=1.25) then you would need to multiply 1.25 by 50 which will get you $62.50 (1.25×50=62.5).

8 0

3 years ago

Help me with this one pls

lora16 [44]

<h3> $= \tt{ - 1.3 - ( - 1.9)}$ </h3><h2> $= \green{\boxed{\tt{{ - 1.3 + 1.9}}}}$ </h2><h3> $= 0.6$ </h3>

4 0

3 years ago

Archaeologists can determine the diets of ancient civilizations by measuring the ratio of carbon-13 to carbon-12 in bones found

Vsevolod [243]

Answer:

Step-by-step explanation:

The null and the alternative hypothesis are:

H₀=μ₀=μ₁=μ₂=μ₃=μ₄

H₁=two or more μ are different X

Let $X_{ij}$ denote the jth observation of the ith sample. The mean and the variance of the ith sample is given by

$Xbar_{i}$ = $\frac{1}{J_{i}} summation(X_{ij})$ where J=1 to $J_{i}$ and $J_{i}$ is ith sample size

$s_{i} ^{2} =\frac{1}{J_{i}-1 }summation (X_{ij} - Xbar_{i})^{2}$

The sample sizes are J₁ =12 J₂=10 J₃=18 J₄=9

the total number in all samples combined is 49

finding Xbar₁ and s₁

Xbar₁= 1÷12(17.2+....+13.4)

Xbar₁= 17.0500

s₁²= 1÷(12-1) [(17.0500-17.2)+...+(17.0500-13.4)]

s₁²=5.1336

Similarly find the means and variances of other samples

$\left[\begin{array}{ccc}i&Mean&variance\\1&17.0500&5.1336\\2&15.4500&13.2317\\3&16.4972&6.9460\\4&15.4444&7.4428\end{array}\right]$

the sample grand mean denoted by Xbar is the average of all sampled items taken together:

Xbar= $\frac{1}{49}$ (17.2+.....+16.7)

Xbar=16.2255

Compute the treatment sum of squares SSTr, the sum of squares SSE and the total sum of squares SST

SSTr = ∑ $J_{i} (Xbar_{i} -Xbar)^{2}$ from i=1 to 49

SSTr= 20.9910

$SSE= \sum\limits^I_{i=1}\sum\limits^{J_i}_{j=1}(Xbar_{ij}-Xbar_i)^{2}$

$SSE=\sum\limits^I_{i=1}(J_i-1)s_i^{2}$

SSE=(12-1)(5.1336)+...(9-1)(7.4428)

SSE=353.1796

SST=SSTr+SSE

SST=374.1706

Find the treatment mean square MSTr and the error mean square MSE:

MSTr= SSTr/(I-1)

MSTr=6.9970

MSE=SSE/(N-I)

MSE=7.8484

$F=\frac{MSTr}{MSE}$

F=0.89

The degrees of freedom are I-1=3 for the numerator and N-I=45 for the denominator. Under H₀, F has an $F_{3,45}$ distribution. To find the P value we consult the F table.

P>0.100

The complete ANOVA table is below

$\left[\begin{array}{cccccc}source&DF&SS&MS&F&P\\Agegroup&3&20.9910&6.9970&0.89&0.453\\Error&45&353.1796&7.8484\\Total&48&374.1706\end{array}\right]$

(b) The P-value is large. A follower of 5% rule would not reject H₀. Therefore, we cannot conclude the concentration ratios differ among the age groups

4 0

4 years ago