All categories

3 years ago

Describe a real-life situation where you would have a remainder

2 answers:

swat323 years ago

8 0

Say you have eight people and nine apples and you want everyone to have the same amount of apples (which would be one), so you split them evenly and you have one apple left.

gavmur [86]3 years ago

6 0

Say you are at a party and there are 37 people and everyone gets the same amount as possible which is 9 per person so you will have 1 left over
HOPE THIS HELPS!!!! ;-)

You might be interested in

PLEASE HELP ASAP!!! WILL MAKE YOU THE BRAINLYIST!!!!

elena55 [62]

15 ppl would fit

Bc if you find the area of the dance floor which is 20 feet times 30 feet it equals 60 feet. Then multiply how much space someone takes which is 2 feet times 2 feet equals 4 feet. Then you divide the area of the dance floor by the number for F space one person takes up which is 60÷4. You would get 15 people

3 0

3 years ago

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

3 years ago

A circle has a radius of 30 cm. What is that circumference of the circle?

Brums [2.3K]

Answer:

C = 60pi cm or approximately 188.4 cm

Step-by-step explanation:

The circumference of a circle is given by

C = 2 * pi *r

C = 2* pi * 30

C = 60 pi

We can approximate pi by 3.14

C is approximately 60*3.14 or 188.4

3 0

4 years ago

Read 2 more answers

The standard deviation of a sample taken from population A is 17.6 for a sample of 25.

lawyer [7]

Answer:

The standard deviation of the sample mean differences is _5.23_

Step-by-step explanation:

We have a sample of a population A and a sample of a population B.

For the sample of population A, the standard deviation $\sigma_A$ is

$\sigma_A = 17.6$

The sample size $n_A$ is:

$n_A = 25$ .

For the sample of population B, the standard deviation $\sigma_B$ is

$\sigma_B = 21.2$

The sample size $n_B$ is:

$n_B = 30$ .

Then the standard deviation for the difference of means has the following form:

$\sigma=\sqrt{\frac{\sigma_A^2}{n_A}+\frac{\sigma_B^2}{n_B}}$

Finally

$\sigma=\sqrt{\frac{17.6^2}{25}+\frac{21.2^2}{30}}\\\\\sigma= 5.23$

7 0

4 years ago

PLS HELP ME

eimsori [14]

Answer:

5.5

Step-by-step explanation:

2,3,3,4,5,7,8,8,8,10,10,12

The mean of this data is 7.5. So what you are trying to figure out is the mean between 2 and 7. Since that is 3 and 4 it would be 3.5. So 3.5 is the Q1. Now you have to figure out the mean between 8 and 12. That is 9. This is Q3. You have to subtract 9-3.5 and that is the interquartile range. So 5.5 is the answer. I hope that made sense.

6 0

2 years ago