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3 years ago

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7500 dollars is placed in an account with an annual interest rate of 5.25%. To the nearest tenth of a year, how long will it tak

e for the account value to reach 34300 dollars?

1 answer:

Margarita [4]3 years ago

3 0

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Your father earned $34,000 per year in 1984. To the nearest dollar, what is that equivalent to in 2014 if the CPI in 2014 is 2

xeze [42]

Answer:

The answer is : 70288.46

Step-by-step explanation:

Your father earned $34,000 per year in 1984.

The CPI in 2014 is 215 and the CPI in 1984 is 104.

The CPI or consumer price index examines the weighted average of prices of goods and services like transportation, food etc.

The Current income equivalent will be = value in 1984 X ( CPI in 2014/CPI in 1984)

$34000\times\frac{215}{104}$ = 70288.46

5 0

3 years ago

You are considering leasing a store site that has 1,500 square feet. The landlord is asking for a rent of $50 per square foot, p

tigry1 [53]

Answer:

$75000/year

$6250/month

Step-by-step explanation:

1500 ft²*$50/(ft²*year) = $75000/year

1 year = 12 month

$75000/12 month= $6250/month

3 0

2 years ago

Please help (if you can’t it’s okay) :>

IrinaVladis [17]

Answer:

the answer is 11%

Step-by-step explanation:

this is because

27/30x100=90

which is 90

but the difference between 100 and 90 is 10

so its 10% but its closer to 11%

so yeah

i am not sure if its correct

3 0

3 years ago

Read 2 more answers

A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and

mr Goodwill [35]

Answer:

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

$\mu = \frac{3X_{1} + 4X_{2} }{8}$

Now

Bias for the estimator = E(μ bar) - μ

= $E( \frac{3X_{1} + 4X_{2} }{8}) - 4.5$

= $\frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5$

= $\frac{3(4.5) + 4(4.5)}{8} - 4.5$

= $\frac{13.5 + 18}{8} - 4.5$

= $\frac{31.5}{8} - 4.5$

= 3.9375 - 4.5

= - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

= Var(μ bar) + [Bias(μ bar, μ)]²

= $Var( \frac{3X_{1} + 4X_{2} }{8}) + 0.3136$

= $\frac{1}{64} Var( {3X_{1} + 4X_{2} }) + 0.3136$

= $\frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})] }) + 0.3136$

= $\frac{1}{64} [{3(57.76) + 4(57.76)}] } + 0.3136$

= $\frac{1}{64} [7(57.76)}] } + 0.3136$

= $\frac{1}{64} [404.32] } + 0.3136$

= $6.3175 + 0.3136$

= 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

6 0

3 years ago

A road is inclined at an angle of 3°. After driving 5330 feet along this road, find the drivers increase in altitude

Wewaii [24]

Hope this helps just some notes

8 0

2 years ago