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Alexxx [7]
3 years ago
15

Two boxes of apples contain a total of 24 apples. If you halve the number of apples in the first box and add 4 apples to the sec

ond box, the total changes to 20 apples. How many apples are in each box initially?
A. 12 and 12

B. 18 and 6

C. 16 and 8

D. 10 and 14

Mathematics
2 answers:
maks197457 [2]3 years ago
7 0

Step-by-step explanation:

let the no.s of apples in the one box be=x

then no.s of apples in the 2nd box =24-x

according to the question,

24-x+4=20

-x= 20-28

-x=-8

x= 8

no.s of apples in one box= 8

and no.s of apples in another box=24-8= 16

option C

Lena [83]3 years ago
5 0

Answer: option C is the correct answer.

Step-by-step explanation:

Let x represent the number of apples in the first box.

Let y represent the number of apples in the second box.

Two boxes of apples contain a total of 24 apples. It means that

x + y = 24

If you halve the number of apples in the first box and add 4 apples to the second box, the total changes to 20 apples. This means that

x/2 + y + 4 = 20

Cross multiplying by 2, it becomes

x + 2y + 8 = 40

x + 2y = 40 - 8

x + 2y = 32- - - - - - - - - - - - - -1

Substituting x = 24 - y into equation 1, it becomes

24 - y + 2y = 32

- y + 2y = 32 - 24

y = 8

x = 24 - y = 24 - 8

x = 16

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4 0
3 years ago
A random sample of 23 items is drawn from a population whose standard deviation is unknown. The sample mean isx⎯ ⎯ x¯ = 840 and
gayaneshka [121]

Answer:

(832.156, \ 847.844)

Step-by-step explanation:

Given data :

Sample standard deviation, s = 15

Sample mean, \overline x = 840

n = 23

a). 98% confidence interval

$\overline x \pm t_{(n-1, \alpha /2)}. \frac{s}{\sqrt{n}}$

$E= t_{( n-1, \alpha/2 )} \frac{s}{\sqrt n}}

$t_{(n-1 , \alpha/2)} \frac{s}{\sqrt n}$

$t_{(n-1, a\pha/2)}=t_{(22,0.01)} = 2.508$

∴ $E = 2.508 \times \frac{15}{\sqrt{23}}$

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So, 98% CI is

$(\overline x - E, \overline x + E)$

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