All categories

2 years ago

Tim has read 10% of a book. He has 45 more pages to finish. How many pages are there in the book?

1 answer:

6 0

if Tim has read 10% of the book, that means he still hasn't read 90% of it yet, and we know that that 90% is 45 pages, so say the book has a total amount of pages of "x", which is 100%.

$\begin{array}{ccll} pages&\%\\ \cline{1-2} 45&90\\ x&100 \end{array}\implies \cfrac{45}{x}=\cfrac{90}{100}\implies \cfrac{45}{x}=\cfrac{9}{10} \\\\\\ 450=9x\implies \cfrac{450}{9}=x\implies 50=x$

You might be interested in

A ten pound bag of cherries cost $33.50. How much is one pound of cherries?

Rashid [163]

Each pound costs $3.35. Hope this helped

3 0

2 years ago

Read 2 more answers

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

3 0

3 years ago

A taxi charges 2.50 plus .50 fee for each mile. If the ride costs 7.50 how many miles did he travel

NeX [460]

Answer: 10 miles

Step-by-step explanation:

Mile 1: 3.00

Mile 2: 3.50

Mile 3: 4.00

Mile 4: 4.50

Mile 5: 5.00

Mile 6: 5.50

Mile 7: 6.00

Mile 8: 6.50

Mile 9: 7.00

Mile 10: 7.50

7 0

2 years ago

Gym A charges $9.00 per hour of childcare. Gym B charges $0.75 per 5 minutes of childcare.

TEA [102]

Answer:

$9.00

Step-by-step explanation:

0.75 times 12 equals 9

3 0

3 years ago

Read 2 more answers

PLEASE HELP!! I'LL GIVE BRAINIEST!! What is the measure of ∠PQR?

My name is Ann [436]

Answer:

Step-by-step explanation:

Trust me

7 0

2 years ago

Read 2 more answers

Tim has read ​10% of a book. He has 45 more pages to finish. How many pages are there in the​ book?

Tim has read 10% of a book. He has 45 more pages to finish. How many pages are there in the book?