In second grade Eliza was given 500 from her grandparents this money was deposited into a savings account that earns 3% annual interest compounded quarterly find the account balance at graduation is there no other deposits or withdrawals in the account

Answers

Answer 1
Answer:

Answer:

account balance at graduation is 674.17

Step-by-step explanation:

given data

deposit money P  =  500

interest rate = 3 % compounded quarterly

we take time period = 10 year ( 2nd grade to 12 grade)

solution

we apply here compound interest formula that is

amount = P * (1+(r)/(4))^(4t)      ..................1

put here value and we get

amount = 500 * (1+(0.03)/(4))^(4* 10)

amount = 674.174

so account balance at graduation is 674.17

Answer 2
Answer:

Answer:

$694.63

Step-by-step explanation:

The formula is p(1+r/n)^tn. P, or the principal amount, is 500 dollars. R, or the rate as a decimal, is 0.03. N, or the number of times compounded yearly, is 4. T, or the time in years, is 11, since there are 11 years from grade 2 to grade 12. 12-2+1=11. Now, plug in the information given. 500(1+0.03/4)^4*11 = 500(1.0075)^44 = 694.63 dollars, rounded to the nearest cent.


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Please answer this correctly without making mistakes

Consider the equation 14*10^0.5w=100 Solve the equation for w. Express the solution as a logarithm in base-10

Answers

Step-by-step explanation:

14*10^0.5w=100

14*10^(0.5w)=100

Solve the equation for w

divide by 14 on both sides

10^(0.5w)=(100)/(14)

Convert exponential form into log form

base is 10 so we convert into log form

log (a)= x  then a=10^x

10^(0.5w)=(100)/(14)

0.5w = log((100)/(14) )\n0.5w = log((50)/(7) )\n\n

divide both sides by 0.5

0.5w = log((50)/(7) )\n\n\nw= 2 log((50)/(7) )

Answer:

w= 2 log((50)/(7) )

There are 45 new houses being built in a neighborhood. Last month, 2/3 of them were sold. This month 2/5, of the remaining houses were sold. How many houses are left to be sold?

Answers

Answer: 9 houses

Step-by-step explanation:

Step 1:

Since 2/3 of the houses were sold last month, 1/3 of them were left.

45 * 1/3 = 15 houses left last month

Step 2:

Since 2/5 of the remaining houses were sold this month, 3/5 of them were left

15 * 3/5 = 9 houses left to be sold

there's your answer hope this helps :)

According to the University of Nevada Center for Logistics Management, 6% of all merchandise sold in the United States gets returned (BusinessWeek, January 1 5, 2007). A Houston department store sampled 80 items sold in January and found that 12 of the items were returned.Construct a point estimate of the proportion of items returned for the population of sales transactions at the Houston store.

Answers

Using the sample proportion, it is found that the point estimate is of 0.15 = 15%.

What is a sample proportion?

A sample proportion is given by the number of desired outcomes divided by the number of total outcomes. It can also serve as the point estimate for the population proportion.

In this problem, 12 out of the 80 items sold were returned, hence:

12/80 = 0.15 = 15%.

The point estimate is of 0.15 = 15%.

More can be learned about point estimates at brainly.com/question/24651197

Answer:

a) \hat p = (X)/(n)= (12)/(80)= 0.15

b) 0.15 - 1.96 \sqrt{(0.15(1-0.15))/(80)}=0.072  

0.15 + 1.96 \sqrt{(0.15(1-0.15))/(80)}=0.228  

And the 95% confidence interval would be given (0.072;0.228).  

c) For this case since the confidence interval not contains the value 0.06 or 6% and since the lower limit for the confidence interval is higher than 0.06 (0.072>0.06), we have enough statistical evidence to support the conclusion that the true proportion of items returned is higher than 0.06 or 6% at a significance of 5%.

Step-by-step explanation:

Assumign the following question for the problem:

a. Construct a point estimate of the proportion of items returned for the population of  sales transactions at the Houston store.

For this case the best estimate for the true proportion is given by the sample proportion:

\hat p = (X)/(n)= (12)/(80)= 0.15

b. Construct a 95% confidence interval for the porportion of returns at the Houston store.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The confidence interval would be given by this formula  

\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_(\alpha/2)=1.96  

And replacing into the confidence interval formula we got:  

0.15 - 1.96 \sqrt{(0.15(1-0.15))/(80)}=0.072  

0.15 + 1.96 \sqrt{(0.15(1-0.15))/(80)}=0.228  

And the 95% confidence interval would be given (0.072;0.228).  

c. Is the proportion of returns at the Houston store significantly different from the returns  for the nation as a whole? Provide statistical support for your answer.

For this case since the confidence interval not contains the value 0.06 or 6% and since the lower limit for the confidence interval is higher than 0.06 (0.072>0.06), we have enough statistical evidence to support the conclusion that the true proportion of items returned is higher than 0.06 or 6% at a significance of 5%.

The expression (x^n) (x^5)^3 is equivalent to x^30. What is the value of n?

Answers

Answer:

n = 15

Step-by-step explanation:

(x^n)(x^5)^3 = x^30

(x^n)(x^15) = x^30

x ^ (n+15) = x^30

n + 15 = 30

n = 15

To do the question, you need to know the exponent rules, which are short cuts to be able to do exponent work.

If you raise a power to a power, then times the exponents.

If you are multiplying two terms with the same base (in this case, x; the base is the large font bottom number) then ADD the exponents.

the cell wall of which out of this is not made up of cellulose bacteria mango tree hydrilla cactus which one is correct​

Answers

Answer:

The cell wall lies outside the plasma membrane. The plant cell wall is mainly composed of cellulose. Cellulose is a complex substance and provides structural strength to plants. Bacteria is not a plant therefore its cell wall is made up of peptidoglycan

Step-by-step explanation:

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The United States Bureau of Labor Statistics (BLS) conducts the Quarterly Census of Employment and Wages (QCEW) and reports a variety of information on each county in America. In the third quarter of 2016, the QCEW reported the total taxable earnings, in millions, of all wage earners in all 3222 counties in America. Suppose that James is an economist who collects a simple random sample of the total taxable earnings of workers in 56 American counties during the third quarter of 2016. According to the QCEW, the true population mean and standard deviation of taxable earnings, in millions of dollars, by county are ?=28.29 and ?=33.493, respectively.
Let X be the total taxable earnings, in millions, of all wage earners in a county. The mean total taxable earnings of all wage earners in a county across all the counties in James' sample is x??.
Use the central limit theorem (CLT) to determine the probability P that the mean taxable wages in James' sample of 56 counties will be less than $33 million. Report your answer to four decimal places.
P(x??<33)=
Use the CLT again to determine the probability that the mean taxable wages in James' sample of 56 counties will be greater than $30 million. Report your answer to four decimal places.
P(x??>30)=

Answers

Answer:

1

  The  probability is  P(\= X < 33) = 0.8531

2

  The  probability is  P(\= X > 30) = 0.3520

Step-by-step explanation:

From the  question we are told that

     The population mean is  \mu =  28.29

      The standard deviation is \sigma  =  33.493

       The sample size is  n  = 56

Generally the standard error for the  sample  mean (\= x ) is mathematically evaluated as

        \sigma _(\=x) =  (\sigma)/(√(n) )

substituting values  

       \sigma _(\=x) =  (33.493)/(√(56) )

      \sigma _(\=x) = 4.48

Apply central limit theorem[CLT] we have  that

        P(\= X < 33) =  [z <  (33 -  \mu )/(\sigma_(\= x)) ]

substituting values

       P(\= X < 33) =  [z <  (33 -  28.29 )/(4.48) ]

       P(\= X < 33) =  [z <  1.05 ]

From the z-table  we have that  

       P(\= X < 33) = 0.8531

For the second question

    Apply central limit theorem[CLT] we have  that

    P(\= X > 30 ) =  [z >  (30 -  \mu )/(\sigma_(\= x)) ]

substituting values

   P(\= X < 33) =  [z >  (30 -  28.29 )/(4.48) ]

From the z-table  we have that  

     P(\= X < 30) = 0.6480

Thus  

     P(\= X > 30) = 1- P(\= X < 30) = 1- 0.6480

     P(\= X > 30) = 0.3520