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3 years ago

Find the volume.

Mathematics

2 answers:

PtichkaEL [24]3 years ago

8 0

Answer:

Step-by-step explanation:

241.428

volume=lwh

=9.3×5.9×4.4

Viktor [21]3 years ago

8 0

<h2>Answer:</h2><h2>241.428 cu.cm</h2>

$solution \\ volume = l\times b \times h \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 9.3 \times 4.4 \times 5.9 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 241.428 \: cu.cm$

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A recent study found that 51 children who watched a commercial for Walker Crisps (potato chips) featuring a long-standing sports

hichkok12 [17]

Answer:

The claim is that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

The kind of test to use is a t -test because a t -test is used to check if there is a difference between means of a population

$t = 3.054$

The p-value is $p-value = P(Z > 3.054) = 0.0011291$

The conclusion is

There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

The test statistics is

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 51$

The first sample mean is $\mu_1 = 36$

The second sample size is $n_2 = 41$

The second sample size is $\mu_2 = 25$

The first standard deviation is $\sigma _1 = 21.4 \ g$

The second standard deviation is $\sigma _2 = 12.8 \ g$

The level of significance is $\alpha = 0.05$

The null hypothesis is $H_o : \mu_1 = \mu_ 2$

The alternative hypothesis is $H_a : \mu_1 > \mu_2$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } }$

=> $t = \frac{ 36 - 25}{ \sqrt{ \frac{ 21.4^2}{51} + \frac{ 12.8^2}{41} } }$

=> $t = 3.054$

The p-value is mathematically represented as

$p-value = P(Z > 3.054)$

Generally from the z table

$P(Z > 3.054) = 0.0011291$

=> $p-value = P(Z > 3.054) = 0.0011291$

From the values obtained we see that $p-value < \alpha$ so the null hypothesis is rejected

Thus the conclusion is

There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

5 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:

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

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

$\mathbf{Total = 302400}$