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dusya [7]
3 years ago
8

Simplifying algebraic expressions W+6W use W=4

Mathematics
1 answer:
nordsb [41]3 years ago
4 0
Hey there Titachely1p09ewg,

Answer:

W + 6W  = 4 + 6(4)
               = 4 + 24
               = 28
Thus, the answer is 28

Hope this helps :D

<em>~Top♥</em>
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solve the system of linear equations using substitution. Which expression would be eastier to substitute into the other equation
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The x=7y-4 one and put it to where the x in the other equation is.
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Identify the median and IQR of the data set below.20222426283032Your answerWhich dota
Dennis_Churaev [7]

We are given a boxplot that contains some data. We will use the information below to analyze the given question.

STEP 1: Median

From the image above, we can see that the median represents the line drawn inside the box

This gives;

\text{Median}=24

STEP 2: Interquartile range

The interquartile range is the difference between the third quartile and the first quartile.

\begin{gathered} q_3=28 \\ q_1=22 \\ \text{IQR}=28-22 \\ \text{IQR}=6 \end{gathered}

4 0
1 year ago
What statement about pawn shops is correct?
aleksklad [387]

Answer:

B. A pawn shop has lower interest rates than banks.

May i please have the brainliest answer?

3 0
4 years ago
Can someone please help me factor 3x^3-19x^2+33x-9
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8 0
3 years ago
A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of
lina2011 [118]

Answer:

We reject H₀, with CI = 90 % we can conclude that the mean numbers of times identified with Ti-84 does not exceed that of the Ti-89 by more than 1,80

Step-by-step explanation:

Ti-89 Calculator

Sample mean     x = 31,5

Sample standard deviation   s₂  = 8,35

Sample size        n₂ = 11

Ti-84 Calculator

Sample mean     y = 46,2

Sample standard deviation   s₁  = 9,99

Sample size        n₁ = 12

t(s)  = ( y - x - d ) / √s₁²/n₁ + s₂²/n₂

t(s)  = 12,9 / √(99,8/12) + (69,72/11)

t(s)  = 12,9 / √8,32 + 6,34

t(s)  = 12,9 / 3,83

t(s) = 3,37

Test Hypothesis

Null Hypothesis               H₀      y - x > 1,80

Altenative Hypothesis       Hₐ      y - x ≤ 1,80

We have a t(s) = 3,37

We need to compare with t(c)   critical value for

t(c) α; n₁ +n₂-2            df = 12 +11 -2     df  = 21

If we choose  CI = 95 %    then  α = 5 %   α = 0,05

From t-student table

t(c) = 1,72

t(s) = 3,37

t(s) >t(c)

t(s) is in the rejection region therefore we accept Hₐ with CI = 95 %

8 0
3 years ago
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