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kirill115 [55]
3 years ago
11

2020-2021 T-Math-Alg1-T1and2-CBT.: Section 2 - Calculator Section Question: 2-4

Mathematics
1 answer:
sammy [17]3 years ago
8 0

Answer:

The correct option is;

1, 6, 11, 15

Step-by-step explanation:

The given information are;

The function that defines the terms of the sequence is f(n) = f(n - 1) + 5

Where;

n = The number of the term

For n = 1, we have, f(1) = -4

The first four terms are therefore;

f(2) = f(2 - 1) + 5 = f(1) + 5 = - 4 + 5 = 1

f(3) = f(3 - 1) + 5 = f(2) + 5 = 1 + 5 = 6

f(4) = f(4 - 1) + 5 = f(3) + 5 = 6 + 5 = 11

f(5) = f(5 - 1) + 5 = f(4) + 5 = 11 + 5 = 15

The first four terms are; 1, 6, 11, 15

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yKpoI14uk [10]
3/4 = (2*3) / (2* 4) = 6/8

7/8.

Is 3/4 greater than 7/8? 
Is like asking is 6/8 greater than 7/8?  Because 3/4 is equivalent to 6/8.

The answer is No.
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3 years ago
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The manufacturer of hardness testing equipment uses​ steel-ball indenters to penetrate metal that is being tested.​ However, the
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Answer:

Check the explanation

Step-by-step explanation:

Let X denotes steel ball and Y denotes diamond

\bar{x_1} = 1/9( 50+57+......+51+53)

=530/9

=58.89

\bar{x_2}= 1/9( 52+ 56+....+ 51+ 56)

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difference = d =(60.33- 58.89)

=1.44

s^2=1/n\sum xi^2 - n/(n-1)\bar{x}^2

s12 = 1/9( 502+572+......+512+532) -9/8 (58.89)2

=31686/8 - 9/8( 3468.03)

=3960.75 - 3901.53

=59.22

s1 = 7.69

s22 = 1/9( 522+ 562+....+ 512+ 562) -9/8 (60.33)2

=33295/8 - 9/8 (3640.11)

=4161.875 - 4095.12

=66.75

s2 =8.17

sample standard deviation for difference is

s=\sqrt{[(n1-1)s_1^2+ (n2-1)s_2^2]/(n1+n2-2)}

 = \sqrt{[(9-1)*59.22+ (9-1)*66.75]/(9+9-2)}

= \sqrt{1007.76/16}

=7.93

sd = s*\sqrt{(1/n1)+(1/n2)}

=7.93*\sqrt{(1/9)+(1/9)}

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=3.74

For 95% confidence level Z (\alpha /2) =1.96

confidence interval is

d\pm Z(\alpha /2)*s_d

=(1.44 - 1.96* 3.75 , 1.44+1.96* 3.75)

=(1.44 - 7.35 , 1.44 + 7.35)

=(-2.31, 8.79)

There is sufficient evidence to conclude that the two indenters produce different hardness readings.

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3 years ago
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