The ratio of boys to girls in Mr. Hakeem's class is 64. If there are 18 boys, how many girls

An author published a book which was being sold online. The first month the author sold 21000 books, but the sales were declinin

Wittaler [7]

Answer: 119,616

You may have to erase the comma when inputting the answer.

=======================================================

Explanation:

The author sells 21,000 books in the first month.

Then they sell 21,000*(1-0.12) = 21,000*(0.88) = 18,480 in the second month.

Then they sell 18,480*(1-0.12) = 18,480*(0.88) = 16,262.4 = 16,262 in the third month.

And so on. These values follow a geometric sequence with first term a = 21,000 and common ratio r = 0.88

We can think of the 0.88 as 88% of the original value (since losing 12%, we keep the remaining 88%)

-------------

Use the formula below to sum the first 9 terms of the geometric sequence

Plug in a = 21000 and r = 0.88

$S_n = a*\frac{1-r^n}{1-r}\\\\S_9 = 21000*\frac{1-0.88^9}{1-0.88}\\\\S_9 \approx 21000*\frac{0.6835216}{0.12}\\\\S_9 \approx 21000*5.6960133\\\\S_9 \approx 119,616.2793\\\\S_9 \approx 119,616\\\\$

The author sold about 119,616 books over the course of the first nine months.

4 0

2 years ago

What is the sale price for your purchase if you have a 20% off coupon and you buy a shirt for $19.99 and pants for $10.

Andrew [12]

Answer: I believe it is $23.99

Explanation:
19.99+10= 29.99
29.99x0.2= 5.992
29.99-5.99= $23.99

I hope this helped!

4 0

3 years ago

Use the quadratic formula to solve 4x^2 - 8x - 60 = 0

zhannawk [14.2K]

A. You can factor it to those terms

4 0

3 years ago

In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt

ankoles [38]

Answer:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

$df=n-1=44-1=43$

$p_v =P(t_{(43)}>1.436)=0.079$

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

$\bar X=13.88$ represent the sample mean

$s=3.14$ represent the sample standard deviation

$n=44$ sample size

$\mu_o =13.2$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:

Null hypothesis: $\mu \leq 13.2$

Alternative hypothesis: $\mu > 13.2$

The statistic for this case is:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculating the statistic

Replacing the info given we got:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

P-value

The degrees of freedom are:

$df=n-1=44-1=43$

Since is a right tailed test the p value is:

$p_v =P(t_{(43)}>1.436)=0.079$

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

8 0

3 years ago

Please can someone help quickly

rodikova [14]

((5-2)*180)/5=SPT
SPT=108
360-(108+90)=RPT
RPT=162
((n-2)*180)/n=162
162n=(n-2)*180
.9n=n-2
.1n=2
n=20

8 0

3 years ago