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Mandarinka [93]
3 years ago
5

Find the sum of the first six terms of the sequence for which a2=0.7 and a3=0.49.

Mathematics
2 answers:
alina1380 [7]3 years ago
8 0
A1 = 0.91
a6 = -0.14
sum = (0.91+-0.14) /2
sum = 2.31
Sloan [31]3 years ago
3 0

Answer:  Sum of the first six terms is 2.31.

Step-by-step explanation:

Since we have given that

a_2=0.7\\\\and\\\\a_3=0.49

Since we know the formula for "Arithmetic Progression":

a_2=a+d=0.7\implies a=0.7-d\\\\a_3=a+2d=0.49

Now, solving the above two equations:

a+2d=0.49\\\\0.7-d+2d=0.49\\\\0.7+d=0.49\\\\d=0.49-0.70\\\\d=-0.21

So, when we put the value of d in the first equation, we get that

a=0.7-d\\\\a=0.7+0.21\\\\a=0.91

so, Sum of first six terms would be

S_6=\dfrac{6}{2}(2\times 0.91+(6-1)\times -0.21)\\\\S_6=3(1.82+5\times -0.21)\\\\S_6=3(1.82-1.05)\\\\S_6=2.31

Hence, Sum of the first six terms is 2.31.

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In a completely randomized experimental design, three brands of paper towels were tested for their ability to absorb water. Equa
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Answer:

Yes. At this significance level, there is evidence to support the claim that there is a difference in the ability of the brands to absorb water.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>The significance level is 0.05.</em>

<em>The data is:</em>

<em>Brand X: 91, 100, 88, 89</em>

<em>Brand Y: 99, 96, 94, 99</em>

<em>Brand Z: 83, 88, 89, 76</em>

We have to check if there is a significant difference between the absorbency rating of each brand.

Null hypothesis: all means are equal

H_0:\mu_x=\mu_y=\mu_z

Alternative hypothesis: the means are not equal

H_a: \mu_x\neq\mu_y\neq\mu_z

We have to apply a one-way ANOVA

We start by calculating the standard deviation for each brand:

s_x^2=30,\,\,s_y^2=6,\,\,s_z^2=35.33

Then, we calculate the mean standard error (MSE):

MSE=(\sum s_i^2)/a=(30+6+35.33)/3=71.33/3=23.78

Now, we calculate the mean square between (MSB), but we previously have to know the sample means and the mean of the sample means:

M_x=92,\,\,M_y=97,\,\,M_z=84\\\\M=(92+97+84)/3=91

The MSB is then:

s^2=\dfrac{\sum(M_i-M)^2}{N-1}\\\\\\s^2=\dfrac{(92-91)^2+(97-91)^2+(84-91)^2}{3-1}\\\\\\s^2=\dfrac{1+36+49}{2}=\dfrac{86}{2}=43\\\\\\\\MSB=ns^2=4*43=172

Now we calculate the F statistic as:

F=MSB/MSE=172/23.78=7.23

The degrees of freedom of the numerator are:

dfn=a-1=3-1=2

The degrees of freedom of the denominator are:

dfd=N-a=3*4-3=12-3=9

The P-value of F=7.23, dfn=2 and dfd=9 is:

P-value=P(F>7.23)=0.01342

As the P-value (0.013) is smaller than the significance level (0.05), the null hypothesis is rejected.

There is evidence to support the claim that there is a difference in the ability of the brands to absorb water.

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Step-by-step explanation:

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Answer: (A) The image of JKL after a 90° counterclockwise about the origin is shown in figure 1. (B) The image of JKL after a reflection across the y-axis is shown in figure 2.

Explanation:

(A)

From the given figure it is noticed that the coordinate points are J(-4,1), K(-4,-2) and L(-3,-1).

If a shape rotate 90 degree counterclockwise about the origin, then,

(x,y)\rightarrow (-y,x)

J(-4,1)\rightarrow J'(-1,-4)

K(-4,-2)\rightarrow K'(2,-4)

L(-3,-1)\rightarrow L'(1,-3)

Therefore, the vertex of imare are  J'(-1,-4),  K'(2,-4) and  L'(1,-3). The graph is shown in figure (1).

(B)

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(x,y)\rightarrow (-x,y)

J(-4,1)\rightarrow J''(4,1)

K(-4,-2)\rightarrow K''(2,-4)

L(-3,-1)\rightarrow L''(3,-1)

Therefore, the vertex of imare are  J''(4,1),  K''(2,-4) and  L''(3,-1). The graph is shown in figure (2).

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