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4 years ago

15

Question 1 (1 point)

1 answer:

sergey [27]4 years ago

4 0

Answer:

10

Step-by-step explanation:

2 1/2 ÷ 1/4 reduce

5/2 × 4/1 = 20/2

20 ÷ 2 = 10 pounds of flour per sugar

(I'm not very good at math so this might be wrong)

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Draw the graph on the following lines.

postnew [5]

Answer:

4. The y would be 9

5. The x-int would be 1.5

Step-by-step explanation:

7 0

3 years ago

The box plots represent the number of minutes costumers spend on hold when calling a company ?

kow [346]

Answer:

8

Step-by-step explanation:

I think it is 8 because I did the equation Q3=3/4(12) and I got 9 which means the answer is 8 which is the 9th number

7 0

3 years ago

Read 2 more answers

Geometric Sequence S = 1.0011892 + ... + 1.0012 + 1.001 + 1

leva [86]

Answer:

<em /> $S_{1893} =5632.98$ <em />

<em />

Step-by-step explanation:

The correct form of the question is:

$S = 1.001^{1892} + ... + 1.001^2 + 1.001 + 1$

Required

Solve for Sum of the sequence

The above sequence represents sum of Geometric Sequence and will be solved using:

$S_n = \frac{a(1 - r^n)}{1 - r}$

But first, we need to get the number of terms in the sequence using:

$T_n = ar^{n-1}$

Where

$a = First\ Term$

$a = 1.001^{1892}$

$r = common\ ratio$

$r = \frac{1}{1.001}$

$T_n = Last\ Term$

$T_n = 1$

So, we have:

$T_n = ar^{n-1}$

$1 = 1.001^{1892} * (\frac{1}{1.001})^{n-1}$

Apply law of indices:

$1 = 1.001^{1892} * (1.001^{-1})^{n-1}$

$1 = 1.001^{1892} * (1.001)^{-n+1}$

Apply law of indices:

$1 = 1.001^{1892-n+1}$

$1 = 1.001^{1892+1-n}$

$1 = 1.001^{1893-n}$

Represent 1 as $1.001^0$

$1.001^0 = 1.001^{1893-n}$

They have the same base:

So, we have

$0 = 1893-n$

Solve for n

$n = 1893$

So, there are 1893 terms in the sequence given.

Solving further:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Where

$a = 1.001^{1892}$

$r = \frac{1}{1.001}$

$n = 1893$

So, we have:

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{1 -\frac{1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{1.001 -1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001^{1893}})}{\frac{0.001}{1.001} }$

Simplify the numerator

$S_{1893} =\frac{1.001^{1892} -\frac{1.001^{1892}}{1.001^{1893}}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{1892-1893}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{-1}}{\frac{0.001}{1.001} }$

$S_{1893} =(1.001^{1892} -1.001^{-1})/({\frac{0.001}{1.001} })$

$S_{1893} =(1.001^{1892} -1.001^{-1})*{\frac{1.001}{0.001}}$

$S_{1893} =\frac{(1.001^{1892} -1.001^{-1}) * 1.001}{0.001}$

Open Bracket

$S_{1893} =\frac{1.001^{1892}* 1.001 -1.001^{-1}* 1.001 }{0.001}$

$S_{1893} =\frac{1.001^{1892+1} -1.001^{-1+1}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1.001^{0}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1}{0.001}$

$S_{1893} =5632.97970294$

Hence, the sum of the sequence is:

<em /> $S_{1893} =5632.98$ <em> ----- approximated</em>

4 0

3 years ago

There are 8 rows and 8 columns, or 64 squares

lawyer [7]

Complete Question:

There are 8 rows and 8 columns, or 64 squares on a chessboard.

Suppose you place 1 penny on Row 1 Column A,

2 pennies on Row 1 Column B,

4 pennies on Row 1 Column C, and so on …

Determine the number of pennies in Row 1

Determine the number of pennies on the entire chessboard?

Answer:

255 in the first row

18,446,744,073,709,551,615 in the entire board

Step-by-step explanation:

Given

$Rows = 8$

$Columns = 8$

Solving (a): Number of pennies in first row

The question is an illustration of geometric sequence which follows

$1,2,4....$

Where

$a =1$ --- The first term

Calculate the common ratio, r

$r = \frac{T_2}{T_1} = \frac{4}{2} = 2$

The number of pennies in the first row will be calculated using sum of n terms of a GP.

$S_n = \frac{a(r^n - 1)}{n - 1}$

Since, the first row has 8 columns, then

$n = 8$

Substitute 8 for n, 2 for r and 1 for a in $S_n = \frac{a(r^n - 1)}{r - 1}$

$S_8 = \frac{1 * (2^8 - 1)}{2 - 1}$

$S_8 = \frac{1 * (256 - 1)}{1}$

$S_8 = \frac{1 * 255}{1}$

$S_8 = 255$

Solving (b): The entire board has 64 cells.

So:

$n = 64$

Substitute 64 for n, 2 for r and 1 for a in $S_n = \frac{a(r^n - 1)}{r - 1}$

$S_{64} = \frac{1 * (2^{64} - 1)}{2 -1}$

$S_{64} = \frac{(2^{64} - 1)}{1}$

$S_{64} = \frac{(18,446,744,073,709,551,616 - 1)}{1}$

$S_{64} = \frac{18,446,744,073,709,551,615}{1}$

$S_{64} = 18,446,744,073,709,551,615$

5 0

3 years ago

Someone pls help pls

muminat

Answer:

Part A: Scale is 0.5

Part B: (4, 0) (3, -1) (-1, 2)

Part C: Yes

Step-by-step explanation:

sorry can't be bothered writing explanation

5 0

3 years ago