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3 years ago

I just need question c

Mathematics

1 answer:

storchak [24]3 years ago

8 0

Answer:

2060.63 m (2061 to the nearest metre).

Step-by-step explanation:

4065^2 = 3504^2 + x^2 where x is the horizontal change.

x^2 = 4065^2 - 3504^2

x^2 = 4246209

x = √(4246209)

= 2060.63 m.

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The digit to the right of 1.249 is

givi [52]

Are you looking for the name of the place value? If you are the answer is ten thousands. If that isn’t what you’re looking for, I need more information.

7 0

3 years ago

2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books

insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

When the books are arranged in any order, the number of arrangements is 3628800
When the mathematics book must not be together, the number of arrangements is 2903040
When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

$\mathbf{Novels = 3}$

$\mathbf{Mathematics = 2}$

$\mathbf{Chemistry = 5}$

(a) The books in any order

First, we calculate the total number of books

$\mathbf{n = Novels + Mathematics + Chemistry}$

$\mathbf{n = 3 + 2 + 5}$

$\mathbf{n = 10}$

The number of arrangement is n!:

So, we have:

$\mathbf{n! = 10!}$

$\mathbf{n! = 3628800}$

(b) The mathematics book, not together

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

$\mathbf{Maths\ together = 2 \times 9!}$

Using the complement rule, we have:

$\mathbf{Maths\ not\ together = Total - Maths\ together}$

This gives

$\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}$

$\mathbf{Maths\ not\ together = 2903040}$

(c) The novels must be together and the chemistry books, together

We have:

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the novels in:

$\mathbf{Novels = 3!\ ways}$

Next, arrange the chemistry books in:

$\mathbf{Chemistry = 5!\ ways}$

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

$\mathbf{n =Mathematics + 1 + 1}$

$\mathbf{n =2 + 1 + 1}$

$\mathbf{n =4}$

So, the number of arrangements is:

$\mathbf{Arrangements = n! \times 3! \times 5!}$

$\mathbf{Arrangements = 4! \times 3! \times 5!}$

$\mathbf{Arrangements = 17280}$

(d) The mathematics must be together and the chemistry books, not together

We have:

$\mathbf{Mathematics = 2}$

$\mathbf{Novels = 3}$

$\mathbf{Chemistry = 5}$

First, arrange the mathematics in:

$\mathbf{Mathematics = 2!}$

Literally, the number of chemistry and mathematics now is:

$\mathbf{n =Chemistry + 1}$

$\mathbf{n =5 + 1}$

$\mathbf{n =6}$

So, the number of arrangements of these books is:

$\mathbf{Arrangements = n! \times 2!}$

$\mathbf{Arrangements = 6! \times 2!}$

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

$\mathbf{Arrangements = ^7P_3}$

So, the total arrangement is:

$\mathbf{Total = 6! \times 2!\times ^7P_3}$

$\mathbf{Total = 6! \times 2!\times 210}$

$\mathbf{Total = 302400}$

Read more about permutations at:

brainly.com/question/1216161

8 0

3 years ago

ASK YOUR TEACHER An article reported that for a sample of 42 kitchens with gas cooking appliances monitored during a one-week pe

xxMikexx [17]

Answer:

$654.16-2.02\frac{165.23}{\sqrt{42}}=602.66$

$654.16+2.02\frac{165.23}{\sqrt{42}}=705.66$

So on this case the 95% confidence interval would be given by (602.66;705.66)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=654.16$ represent the sample mean

$\mu$ population mean (variable of interest)

s=165.23 represent the sample standard deviation

n=42 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=42-1=41$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,41)".And we see that $t_{\alpha/2}=2.02$

Now we have everything in order to replace into formula (1):

$654.16-2.02\frac{165.23}{\sqrt{42}}=602.66$

$654.16+2.02\frac{165.23}{\sqrt{42}}=705.66$

So on this case the 95% confidence interval would be given by (602.66;705.66)

7 0

4 years ago

Anthony buys a bag of oranges for $5.79. He pays with a $10 bill. How much change does Anthony receive? Enter your answer in the

Nata [24]

Answer: I don’t know if you’re being serious or what but $4.21

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The volume of a sphere with radius r is 4 3 πr 3 . what is the volume of a sphere with radius 5

MA_775_DIABLO [31]

Sphere volume = (4/3) PI * radius ^ 3
sphere volume = (4/3) PI * 5 ^ 3
sphere volume = (4/3) PI * 125
sphere volume = 523.5987755983 
sphere volume = 523.6 (rounded)

3 0

3 years ago