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3 years ago

If a starting number is 30 and it falls to 11 what % decrease is that?

Mathematics

1 answer:

jonny [76]3 years ago

3 0

Answer:

The question would've given you to what number you should round up your answer.

Answer is:

Whole number:

63%

1 decimal place:

63.3%

2 decimal places:

63.33%

Exact number:

$\frac{190}{3}$ %

Step-by-step explanation:

So the starting no. = 30

Ending no. = 11

In order to evaluate the percentage of decrease we:

(Ending number/starting number) * 100

So:

$\frac{11}{30} *100 = \frac{11}{3} *10 = \frac{110}{3}%$ %

To find the Decrease percentage:

100 % - 110/3 %

= 63.333%

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Alexus [3.1K]

Answer:

a) the common difference is 20

b) $x_8=115 , x_{12}=195$

c) the common difference is -13

d) $a_{12}=52, a_{15}=13$

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

36-16 =20

So, the common difference is 20

b) what is x_8? what is x_12

The formula used is: $x_n=x_1+(n-1)d$

We know common difference d= 20, we need to find $x_1$

Using $x_3=16$ we can find $x_1$

$x_n=x_1+(n-1)d\\x_3=x_1+(3-1)d\\15=x_1+2(20)\\15=x_1+40\\x_1=15-40\\x_1=-25$

So, We have $x_1 = -25$

Now finding $x_8$

$x_n=x_1+(n-1)d\\x_8=x_1+(8-1)d\\x_8=-25+7(20)\\x_8=-25+140\\x_8=115$

So, $\mathbf{x_8=115}$

Now finding $x_{12}$

$x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195$

So, $\mathbf{x_{12}=195}$

c) what is the common difference of the sequence $a_m$

Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

91-104 =-13

So, the common difference is -13

d) what is a_12? what is a_15?

The formula used is: $a_n=a_1+(n-1)d$

We know common difference d= -13, we need to find $a_1$

Using $a_7=104$ we can find $x_1$

$a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195$

So, We have $a_1 = 195$

Now finding $a_{12}$ , put n=12

$a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52$

So, $\mathbf{a_{12}=52}$

Now finding $a_{15}$ , put n=15

$a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13$

So, $\mathbf{a_{15}=13}$

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Step-by-step explanation:

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