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SashulF [63]
3 years ago
10

Dasha has 4 times more rose bushes in her garden than Anna. After 39 rose bushes were replanted into Anna's garden, the number o

f bushes in both gardens became the same. How many rose bushes did Dasha's have originally?
Mathematics
2 answers:
Gnom [1K]3 years ago
6 0

Hi There!

-----------------------------------------------------------

Solution:

a = anna

d = dasha

a + 39 = d * 4

We can say that:

a = 1

d = 10

1 + 39 = 10 * 4

40 = 40

-----------------------------------------------------------

Answer:

Dasha had 10 rose bushes originally.

-----------------------------------------------------------

I hope this helps :)

schepotkina [342]3 years ago
3 0

Answer:

104

Step-by-step explanation:

Hello! :)

-----------------------------------------------------------------------------------------------------------------

Dasha=d, Anna=a

d=4a

d-39=a+39

d=a+78

Plug in d as 4a:

4a=a+78

Solve for a:

3a=78

a=26

Because d=4a, plug in a:

d=4*26

d=104

Dasha originally have 104 rose bushes

-----------------------------------------------------------------------------------------------------------------

I hope this helps :)

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Step-by-step explanation:

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Answer:

Check the explanation

Step-by-step explanation:

Let X denotes steel ball and Y denotes diamond

\bar{x_1} = 1/9( 50+57+......+51+53)

=530/9

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\bar{x_2}= 1/9( 52+ 56+....+ 51+ 56)

=543/9

=60.33

difference = d =(60.33- 58.89)

=1.44

s^2=1/n\sum xi^2 - n/(n-1)\bar{x}^2

s12 = 1/9( 502+572+......+512+532) -9/8 (58.89)2

=31686/8 - 9/8( 3468.03)

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s1 = 7.69

s22 = 1/9( 522+ 562+....+ 512+ 562) -9/8 (60.33)2

=33295/8 - 9/8 (3640.11)

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sample standard deviation for difference is

s=\sqrt{[(n1-1)s_1^2+ (n2-1)s_2^2]/(n1+n2-2)}

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= \sqrt{1007.76/16}

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sd = s*\sqrt{(1/n1)+(1/n2)}

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=7.93* 0.47

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For 95% confidence level Z (\alpha /2) =1.96

confidence interval is

d\pm Z(\alpha /2)*s_d

=(1.44 - 1.96* 3.75 , 1.44+1.96* 3.75)

=(1.44 - 7.35 , 1.44 + 7.35)

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There is sufficient evidence to conclude that the two indenters produce different hardness readings.

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