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const2013 [10]
2 years ago
11

Oleg, Sasha, and Dima share 600 toys. Sasha has twice as many toys as Oleg. Dima has 40 more toys than Oleg. How many toys does

Oleg have?
Mathematics
1 answer:
Anika [276]2 years ago
6 0

Answer: 140

Oleg, Sasha, and Dima shared 600 toys. Sasha had twice as many toys as Oleg. Dima had 40 more toys than Oleg.

Let the number of toys Oleg has = x

Let the number of toys Sasha has = y

Let the number of toys Dima has = z

Sasha had twice as many toys as Oleg

So y = 2*x= 2x

Dima had 40 more toys than Oleg.

z = x + 40

Oleg, Sasha, and Dima shared 600 toys

x + y + z= 600

We know y=2x and z=x+40

Replace it for y and z

Subtract 40 on both sides

Divide by 4 on both sides

Oleg had 140 toys

Step-by-step explanation:

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<span>5 - 4 + 7x + 1 = 7 x + 1</span>
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At the Foremost State Bank the average savings account balance in 2012 was $1900. A random sample of 45 savings account balanes
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The calculated  |t| = |-26.517|=26.517 > 2.326 at 0.01 level of significance

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Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Mean of the Population = 1900

Given random sample size 'n' =  45

Mean of the sample x⁻ = 1000

Standard deviation of the sample = 225

<u><em>Null Hypothesis</em></u>:-

There is no difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012.

<u><em>Alternative Hypothesis</em></u> :-

There is difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012

<u><em>Step(ii</em></u>):-

Test statistic

        t = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }

        t = \frac{1000-1900 }{\frac{225}{\sqrt{45} } }

       t = -900 /33.54 = -26.517

      |t| = |-26.517|=26.517

Degrees of freedom

                       γ = n-1 = 45-1 =44

t₍₀.₀₁ , ₄₄₎ = 2.326

The calculated  |t| = |-26.517|=26.517 > 2.326 at 0.01 level of significance

Null hypothesis is rejected at 0.01 level of significance

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Answer:

= 56.7 miles per hour

Step-by-step explanation:

Step-by-step explanation:

To find the average speed for the entire journey, we need to find the entire distance   and the  total time

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Rounding to the nearest tenth

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