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Tju [1.3M]
3 years ago
6

A collection of 8 backpacks has a mean weight of 14 pounds. A different collection of 12 backpacks has a mean weight of 9 pounds

. What is the mean weight of the 20 backpacks?
Mathematics
2 answers:
Setler79 [48]3 years ago
7 0

Answer:

11

Step-by-step explanation:

To calculate total weight of 8 backpacks. You multiply 8*14=112

To calculate total weight of 12 backpacks is 108 pounds. 12*9=108

Next, you have to calculate the total weight of the whole 20 backpacks 112+108=220

Lastly, you find the mean weight of 20 backpacks. So you divide 220 by 20.

Answer: 11 pounds

Hope this helped!

g100num [7]3 years ago
3 0

Answer:

0.1235

Step-by-step explanation:

Divide 14 and 8 and also divide 12and 9 then u take those two answers and add them up and you divide that answer by20

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Your beginning balance on your lunch account is $42. You buy lunch for $1.80 every day and sometimes buy a snack for $0.85. Afte
tangare [24]

Answer:

The answer to your question is: I bought 7 snacks

Step-by-step explanation:

Data

beginning balance = $42 = b

lunch = $1.80  = l

snack = $ 0.85  = s

final balance = $0.05 = f

                        f = b - 1.8l - 0.05s

                     0.05 = 42 - 1.8l - 0.85s

After 20 days I spent = 1.8(20) in lunches = $36

                    0.05 = 42 - 36 - 0.85s

                   0.05 = 6 - 0.85s

                  0.05 - 6 = -0.85s

                  -5.95 = -0.85s

                  s = -5.95/-0.85

                  s = 7

         

4 0
3 years ago
Three times a larger number is 30 more than 5 times a smaller number. The sum of the larger number and 5 times the smaller numbe
slava [35]
3y  = 5x + 30

y + 5x = 50 .......(1)

3y - 5x = 30 .....(2)        - rearranging the first equation.

Add (1) and (2):-

4y = 80

y = 20  
Now plug y = 20 into equation (1):-
20 + 5x = 50
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The 2 numbers are 6 and 20.

5 0
3 years ago
Read 2 more answers
How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?
Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+...  \ \ \  \ \ --- (1)

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)          

f'(x) = \dfrac{1}{1+x}

f'(0)   = \dfrac{1}{1+0}=1

f ' ' (x)    = \dfrac{1}{(1+x)^2}

f ' ' (x)   = \dfrac{1}{(1+0)^2}=-1

f '  ' '(x)   = \dfrac{2}{(1+x)^3}

f '  ' '(x)    = \dfrac{2}{(1+0)^3} = 2

f ' '  ' '(x)    = \dfrac{6}{(1+x)^4}

f ' '  ' '(x)   = \dfrac{6}{(1+0)^4}=-6

f ' ' ' ' ' (x)    = \dfrac{24}{(1+x)^5} = 24

f ' ' ' ' ' (x)    = \dfrac{24}{(1+0)^5} = 24

Now, the next process is to substitute the above values back into equation (1)

f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \  '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...

In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...

In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...

To estimate the value of In(1.4), let's replace x with 0.4

In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...

In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...

Therefore, from the above calculations, we will realize that the value of \dfrac{0.4^5}{5}= 0.002048 as well as \dfrac{0.4^6}{6}= 0.00068267 which are less than 0.001

Hence, the estimate of In(1.4) to the term is \dfrac{0.4^5}{5} is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0
3 years ago
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