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Bingel [31]
2 years ago
11

If the inflation rate is positive, purchasing power . This situation is reflected in the rate of return on an investment, which

will be the rate of return.
Mathematics
1 answer:
kirill [66]2 years ago
6 0

Answer:

If inflation rate is positive, purchasing power of money decreases thus the rate of returns on investment decreases.

Step-by-step explanation:

When the prices of goods and services are averagely increasing we say that inflation is positive.Purchasing power is high when a currency value is able to purchase more goods and services as per unit of money can buy.when inflation is positive, the amounts of goods and services that can be purchased is reduced.

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Andrew invested $20,000 in mutual funds and received a sum of $35,000 at the end of the investment period. Calculate the ROI.
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Answer: $15,000

As $35,000-$20,000=$15,000.

That's basically the profit he had made, or the return on his investment (ROI).

As he already had $20,000, he had "only" made $15,000.
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A right triangle has side lengths AC = 7 inches, BC = 24 inches, and AB = 25 inches.
rusak2 [61]

<span>m∠A =73.7° 
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m∠C = 90<span>°</span></span>
3 0
2 years ago
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Mr. Washington is putting his DVDs on a shelf that is 10 2⁄3 inches long. If each DVD is 11⁄20 inches wide, how many DVDs can he
omeli [17]
(10 2/3)/(11/20)

(32/3)/(11/20)

(32/3)*(20/11)

640/33

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2 years ago
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Traffic speed: The mean speed for a sample of cars at a certain intersection was kilometers per hour with a standard deviation o
aliina [53]

Answer:

Step-by-step explanation:

Hello!

X₁: speed of a motorcycle at a certain intersection.

n₁= 135

X[bar]₁= 33.99 km/h

S₁= 4.02 km/h

X₂: speed of a car at a certain intersection.

n₂= 42 cars

X[bar]₂= 26.56 km/h

S₂= 2.45 km/h

Assuming

X₁~N(μ₁; σ₁²)

X₂~N(μ₂; σ₂²)

and σ₁² = σ₂²

<em>A 90% confidence interval for the difference between the mean speeds, in kilometers per hour, of motorcycles and cars at this intersection is ________.</em>

The parameter of interest is μ₁-μ₂

(X[bar]₁-X[bar]₂)±t_{n_1+n_2-2} * Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }

t_{n_1+n_2-2;1-\alpha /2}= t_{175; 0.95}= 1.654

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{134*16.1604+41*6.0025}{135+42-2} } = 3.71

[(33.99-26.56) ± 1.654 *(3.71*\sqrt{\frac{1}{135} +\frac{1}{42} })]

[6.345; 8.514]= [6.35; 8.51]km/h

<em>Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15</em>

t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06

[(475.12-321.34) ± 2.485 *(33.06*\sqrt{\frac{1}{12} +\frac{1}{15} })]

[121.96; 185.60]

I hope this helps!

3 0
2 years ago
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