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wlad13 [49]
1 year ago
12

Shawna has $750 in the bank.she deposits $37.50 each week.Ruben has $850 in the bank.He deposits his paycheck of $102.75 every M

onday,and he spends about $70.25 each week.After how many weeks will they have the same amount of money in the bank?
Mathematics
1 answer:
erastova [34]1 year ago
8 0

Answer:

Step-by-step explanation:

Answer: it would take 20 weeks before the amount in both accounts would be the same.

Step-by-step explanation:

Let x represent the number of weeks that it will take either Ruben and Shawna to have the same amount of money in their account.

Let y represent the total amount that would be in Shawna's account after x weeks

Let z represent the total amount that would be in Ruben's account after x weeks

Shawna has $750 in the bank. She deposits $37.50 each week. This means that the total amount after x weeks would be

y = 37.5x + 750

Ruben has $850 in the bank. He deposits his paycheck of $102.75 every Monday,and he spends about $70.25 each week.. This means that the total amount after x weeks would be

z = 850 + 102.75x - 70.25x

z = 850 + 32.5x

To determine the number of weeks before the amount in both accounts will becomes the same, we would equate y to z. It becomes

37.5x + 750= 850 + 32.5x

37.5x - 32.5x = 850 - 750

5x = 100

x = 100/5 = 20

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Answer:

\large \boxed{\text{3 apples and 6 bananas}}

Step-by-step explanation:

 Let a = the number of apples

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Check:

\begin{array}{cccl}3(1.75) + 6(0.40) = 7.65 & \qquad & 3 + 6 = 9\\5.25 + 2.40 = 7.65 & \qquad & 9 = 9\\7.65 = 7.65 & \qquad & \\\end{array}

OK.

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Answer:

<em>a)95%  confidence intervals for the population mean of light bulbs in this batch</em>

(325.5 ,374.5)

b)

<em>The calculated value Z = 4 > 1.96 at 0.05 level of significance</em>

<em>Null hypothesis is rejected </em>

<em>The manufacturer has not right to take the average life of the light bulbs is 400 hours.</em>

Step-by-step explanation:

Given sample size n = 64

Given  mean of the sample x⁻ = 350

Standard deviation of the Population σ = 100 hours

The tabulated value Z₀.₉₅ = 1.96

<em>95%  confidence intervals for the population mean of light bulbs in this batch</em>

<em></em>(x^{-} - Z_{\frac{\alpha }{2} } \frac{S.D}{\sqrt{n} } , x^{-} + Z_{\frac{\alpha }{2} }\frac{S.D}{\sqrt{n} } )<em></em>

<em></em>(350 - 1.96\frac{100}{\sqrt{64} } , 350 + 1.96\frac{100}{\sqrt{64} } )<em></em>

(350 -24.5, 350 +24.5)

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b)

<u><em>Explanation</em></u>:-

Given mean of the Population μ = 400

Given sample size n = 64

Given  mean of the sample x⁻ = 350

Standard deviation of the Population σ = 100 hours

<u><em>Null hypothesis</em></u> : H₀:The manufacturer has right to take the average life of the light bulbs is 400 hours.

μ = 400

<u><em>Alternative Hypothesis: H₁:</em></u> μ ≠400

<u><em>The test statistic </em></u>

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

Z = \frac{350 -400}{\frac{100}{\sqrt{64} } }

|Z| = |-4|

The tabulated value   Z₀.₉₅ = 1.96

The calculated value Z = 4 > 1.96 at 0.05 level of significance

Null hypothesis is rejected.

<u><em>Conclusion:</em></u>-

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