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3 years ago

Need help please will mark brainliest

Mathematics

1 answer:

Luda [366]3 years ago

6 0

Step-by-step explanation:

Maximum = 62

Median = (34+37+39+32+48+45+53+62+58+61+60+41)/12= 47.5≈48

quartile

In increasing order

32, 34, 37, 39, 41, 45, 48, 53, 58, 60, 61, 62

Upper quartile= (58+60)/2 = 59

Lower quartile= (37+39)/2 = 38

Minimum= 32

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The answer is 5/21

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Given f(x)=-5x+1 what is f(-3) a.9 b.-14 c.19 d.16

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<h3>Answer:</h3>

b= -14

Step-by-step explanation:

f (x)= -5x + 1

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Here's a hint on how to add negative# and positive numbers:

Same team ...Yay its a positive

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2+7=9

Opposite team ... Nay its a negative

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3 years ago

5. Find the first, fourth, and tenth terms of the arithmetic

Lubov Fominskaja [6]

The first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

Explanation:

The given rule for the arithmetic sequence is $A(n)=-6+(n-1)(\frac{1}{5} )$

We need to determine the first, fourth and tenth terms of the sequence.

To find the first, fourth and tenth terms, let us substitute $n=1,4,10$ in the general rule for the arithmetic sequence.

To find the first term, substitute $n=1$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(1)=-6+(1-1)(\frac{1}{5} )$

$A(1)=-6+(0)(\frac{1}{5} )$

$A(1)=-6$

Thus, the first term of the arithmetic sequence is -6.

To find the fourth term, substitute $n=4$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(2)=-6+(4-1)(\frac{1}{5} )$

$A(2)=-6+(3)(\frac{1}{5} )$

$A(2)=\frac{-30+3}{5}$

$A(2)=\frac{-27}{5}$

Thus, the fourth term of the arithmetic sequence is $-\frac{27}{5}$

To find the tenth term, substitute $n=10$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(10)=-6+(10-1)(\frac{1}{5} )$

$A(10)=-6+(9)(\frac{1}{5} )$

$A(10)=-6+\frac{9}{5}$

$A(10)=-\frac{21}{5}$

Thus, the tenth term of the arithmetic sequence is $-\frac{21}{5}$

Hence, the first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

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