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Arturiano [62]
1 year ago
6

If a certain number is increased by 5 one half of the result is three-fifths of the excess of 61 over the number find the number

​
Mathematics
1 answer:
PSYCHO15rus [73]1 year ago
8 0

Answer:

= 391/11

Let say number = N

a certain number is increased by 5

= N + 5

,one-half of the result

= (N + 5)/2

Three -fifths of the excess of 61 over the number.

= (3/5)(61 - N)

Equating both

(N + 5)/2  = (3/5)(61 - N)

multiplying by 10 both sides

=> 5(N + 5) = 6(61 - N)

=> 5N + 25 = 366 - 6N

=> 11N = 391

=> N = 391/11

Step-by-step explanation:

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300

Step-by-step explanation:

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1 year ago
Resale Value Garland Mills purchased a certain piece of machinery 2 years ago for $500,000. Its present resale value is $420,000
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Answer:

260,000

Step-by-step explanation:

The value decreases by 40,000 per year (500000-42000/2 years) the value will decrease by 240000 in six (4 years plus 2 years already passed) years. 6 x 40,000 =240000

So, the value should be 260,000 (500000-240000)

Or

year 1 500000-40000=460000

year 2 460000-40000=420000

year 3 420000-40000=380000

year 4 380000-40000=340000

year 5 340000-40000=300000

year 6 300000-40000=260000

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1 year ago
the richmond park gamekeeper wanted to know how many deer were in the park . he caught 90 of them and put a small tag round a ho
Alenkasestr [34]

Answer:

a. Since 10 out of 70 had a tag we can write the ratio 10:70. We need to solve for x in 90:x. x = 630 so the answer is 630.

b. An assumption he could have made is that the number of deer that had a tag and the total amount of deer were directly proportional.

5 0
1 year ago
The quality control manager at a light bulb factory needs to estimate the mean life of a batch (population) of light bulbs. We a
Nadusha1986 [10]

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)

(325.5 ,374.5)

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>-

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

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