All categories

1 year ago

If £1 is equal to 99.72 rupees and the post office only has 500 rupee papers how much can I change out of £450

1 answer:

7 0

£1 = 99.72 rupees -> £450 = 44874 rupees.

As the post office only has 500 rupee papers, once we calculate out the result, we have to round down to the nearest whole number.

The amount of rupee papers we can change out is : 44874 : 500 = 89.748 (papers)
We round down to 89 papers.

You might be interested in

1 2/5 multiplied by 3 4/5

tekilochka [14]

1 2/5 = 7/5
3 4/5 = 19/5
7/5 * 19/5 = 133/25 = 5.32

8 0

3 years ago

Read 2 more answers

What is a nations gross domestic product ? HELP

Vesnalui [34]

the value of goods and services a nation provides in one year

7 0

3 years ago

the flagpole in front of silver pine elementary school is 18.374 feet tall what is 18375 roundef to the nearest tenth

MAXImum [283]

18.374 rounded to the nearest tenth is 18.400.

Because 7 is after 5 so we will round to the next tenth......18.400

Hope I helped:P

3 0

3 years ago

Read 2 more answers

A line passes through the points (9, 30) and (18, 30). Which statement is true about the line?

Arisa [49]

(9,30)(18,30)....when u have 2 sets of points and there y value is the same, then u have a horizontal line with a slope of 0

answer : A...it has a slope of 0 because the change in the y values

7 0

4 years ago

Read 2 more answers

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago