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Bess [88]
3 years ago
8

Question 1

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
5 0

Answer:

You'll need to look at the attached formula.

Annual Payout = 500,000 * .12 * (1.12)^15 / [(1.12)^15  -1]

Annual Payout = 500,000 * .12 * 5.4735657593 / [ 5.4735657593 -1]

Annual Payout = 500,000 * .12 * 5.4735657593 / [4.4735657593]

Annual Payout = 60,000 * 5.4735657593 / [4.4735657593]

Annual Payout = 73,412.12

Source: http://www.1728.org/annupay.htm

That's a LOT of calculating :-)

Step-by-step explanation:

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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

\hat p=\frac{X}{n}=\frac{224}{269}=0.8327

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

        =0.1673

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

  =1-\frac{X}{n}

  =1-\frac{546}{709}

  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

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

     =0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0
3 years ago
Plz plz help<br>I've been stuck on this since yesterday
Vanyuwa [196]

y=ax+b

then

-9= -9a+b

0= -6a+b


a=3 and b=18


y=3x+18

7 0
3 years ago
Read 2 more answers
Josie is training for a race.The ratio of the number of minutes she runs to the number of miles she runs is 24 to 3 she plans to
Fiesta28 [93]

Answer:

80 minutes

Step-by-step explanation:

She runs an 8 minute mile (which you can do by dividing 24/3), then how long will it take her to run 10 miles?

All you have to do from there is multiply 8x10, which gives you 80.

It'll take 80 minutes of 1:20 hours

7 0
2 years ago
Read 2 more answers
What are the solutions of the equation y = negative 2 x squared + 9 x minus 4 shown in the graph below?
aleksandr82 [10.1K]

Answer:the answer c

Step-by-step explanation:

3 0
2 years ago
How many sets of two or more consecutive positive integers can be added to obtain a sum of 1800?
Nataliya [291]

Answer:

n = 60

Step-by-step explanation:

GIVEN DATA:

Total sum of consecutive number is 1800

sum of n number is given as

sum = \frac{ n(n+1)}{2}

where n is positive number and belong to natural number i.e 1,2,3,4,...

from the data given we have1800 = \frac{n(n+1)}{2}

solving for n we get

[/tex]n^2 + n -3600 = 0[/tex]

n = 59.5, -60.5

therefore n = 60

4 0
2 years ago
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