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3 years ago

When is estimation and affective way to determine an answer

Mathematics

1 answer:

Readme [11.4K]3 years ago

5 0

With pi, repeating decimals, etc. Hope this helped!

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A farmer has a basket of peaches. He gives ⅓ of the peaches to one person, ¼ to another, ⅕ to another, ⅛ to another, and then gi

inessss [21]

Answer:

$\frac{1429}{120}$ or $11\frac{109}{120}$

Step-by-step explanation:

Given:

A farmer has a basket of peaches. He gives ⅓ of the peaches to one person, ¼ to another, ⅕ to another, ⅛ to another, and then gives 7 peaches to a 5th person.

Remaining peaches = 4

We need to find the original number of peaches in the basket.

The farmer gives the total number of peaches = $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{8}+7$

Let x be the former gives the total number of peaches

We multiply and divide by 120 in right side of the above equation because of 120 is divided by all given denominator and then simplify.

$x = \frac{120}{120}(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{8}+7)$

$x = \frac{1}{120}(\frac{120}{3}+\frac{120}{4}+\frac{120}{5}+\frac{120}{8}+7\times 120)$

$x = \frac{1}{120}(40+30+24+15+840)$

$x = \frac{1}{120}(949)$

$x = \frac{949}{120}$

We add the remaining peaches by given peaches for the original number of peaches in the basket.

Original number of peaches = $\frac{949}{120}+4$

Original number of peaches = $\frac{949+4\times 120}{120}$

Original number of peaches = $\frac{949+480}{120}$

Original number of peaches = $\frac{1429}{120}$

Original number of peaches = $11\frac{109}{120}$

Therefore the original number of peaches in the basket is $\frac{1429}{120}$ or $11\frac{109}{120}$

5 0

3 years ago

What is the equation of this line in standard form

Anestetic [448]

Answer:

6x-11y=-13

Step-by-step explanation:

(x1,y1)=(3/2,2) and (x2,y2)=(-4,-1)

y-y1= y2-y1/x2-x1 (x-x1)

y-2=-3/-11/2(x-3/2)

=6/11(x-3/2)

11(y-2)=6x-9

11y-22=6x-9

6x-11y=9-22

6x-11y=-13

7 0

3 years ago

Jc bent 26.32 on 8 gallons of gas how much wood he spend for 20

dedylja [7]

1) divide 26.32 by 8 = 3.29
2) 3.29 x 20= 65.80
3) he would spend 65.80 on 20 gallons of gas

4 0

3 years ago

If the test for a disease is accurate 45% of the time, How often will it come back negative if a patient has the disease?

Dafna1 [17]

D. 55% of the time

45% of the time it will say positive

The other 55% of the time will come back negative.

7 0

3 years ago

Activity 4: Performance Task

Nookie1986 [14]

An arithmetic progression is simply a progression with a common difference among consecutive terms.

The sum of multiplies of 6 between 8 and 70 is 390
The sum of multiplies of 5 between 12 and 92 is 840
The sum of multiplies of 3 between 1 and 50 is 408
The sum of multiplies of 11 between 10 and 122 is 726
The sum of multiplies of 9 between 25 and 100 is 567
The sum of the first 20 terms is 630
The sum of the first 15 terms is 480
The sum of the first 32 terms is 3136
The sum of the first 27 terms is -486
The sum of the first 51 terms is 2193

(a) Sum of multiples of 6, between 8 and 70

There are 10 multiples of 6 between 8 and 70, and the first of them is 12.

This means that:

$\mathbf{a = 12}$

$\mathbf{n = 10}$

$\mathbf{d = 6}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{10} = \frac{10}2(2*12 + (10 - 1)6)}$

$\mathbf{S_{10} = 390}$

(b) Multiples of 5 between 12 and 92

There are 16 multiples of 5 between 12 and 92, and the first of them is 15.

This means that:

$\mathbf{a = 15}$

$\mathbf{n = 16}$

$\mathbf{d = 5}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*15 + (16 - 1)5)}$

$\mathbf{S_{16} = 840}$

(c) Multiples of 3 between 1 and 50

There are 16 multiples of 3 between 1 and 50, and the first of them is 3.

This means that:

$\mathbf{a = 3}$

$\mathbf{n = 16}$

$\mathbf{d = 3}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*3 + (16 - 1)3)}$

$\mathbf{S_{16} = 408}$

(d) Multiples of 11 between 10 and 122

There are 11 multiples of 11 between 10 and 122, and the first of them is 11.

This means that:

$\mathbf{a = 11}$

$\mathbf{n = 11}$

$\mathbf{d = 11}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{11}2(2*11 + (11 - 1)11)}$

$\mathbf{S_{11} = 726}$

(e) Multiples of 9 between 25 and 100

There are 9 multiples of 9 between 25 and 100, and the first of them is 27.

This means that:

$\mathbf{a = 27}$

$\mathbf{n = 9}$

$\mathbf{d = 9}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{9} = \frac{9}2(2*27 + (9 - 1)9)}$

$\mathbf{S_{9} = 567}$

(f) Sum of first 20 terms

The given parameters are:

$\mathbf{a = 3}$

$\mathbf{d = 3}$

$\mathbf{n = 20}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{20} = \frac{20}2(2*3 + (20 - 1)3)}$

$\mathbf{S_{20} = 630}$

(f) Sum of first 15 terms

The given parameters are:

$\mathbf{a = 4}$

$\mathbf{d = 4}$

$\mathbf{n = 15}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{15} = \frac{15}2(2*4 + (15 - 1)4)}$

$\mathbf{S_{15} = 480}$

(g) Sum of first 32 terms

The given parameters are:

$\mathbf{a = 5}$

$\mathbf{d = 6}$

$\mathbf{n = 32}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{32} = \frac{32}2(2*5 + (32 - 1)6)}$

$\mathbf{S_{32} = 3136}$

(g) Sum of first 27 terms

The given parameters are:

$\mathbf{a = 8}$

$\mathbf{d = -2}$

$\mathbf{n = 27}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{27} = \frac{27}2(2*8 + (27 - 1)*-2)}$

$\mathbf{S_{27} = -486}$

(h) Sum of first 51 terms

The given parameters are:

$\mathbf{a = -7}$

$\mathbf{d = 2}$

$\mathbf{n = 51}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{51} = \frac{51}2(2*-7 + (51 - 1)*2)}$

$\mathbf{S_{51} = 2193}$

Read more about arithmetic progressions at:

brainly.com/question/13989292

4 0

2 years ago

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