All categories

3 years ago

Find the surface area of the cube with length of a side s=9.6 in.

Mathematics

2 answers:

g100num [7]3 years ago

4 0

Cube it. So: 9.6 cubed

Allisa [31]3 years ago

3 0

Answer:

the surface area= 6s^2 = 6(9.6)^2 = 552.96 in^2

Step-by-step explanation:

You might be interested in

Tres expresiones fraccionales equivalentes a 25%

Oksana_A [137]

Answer:

1/4

Step-by-step explanation:

$25\% = \frac{25}{100} = \frac{1}{4} \\$

8 0

3 years ago

A film distribution manager calculates that 7% of the films released are flops. If the manager is correct, what is the probabili

diamong [38]

Answer:

0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.07, n = 404$ . So

$\mu = E(X) = np = 404*0.07 = 28.28$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{404*0.07*0.93} = 5.13$

If the manager is correct, what is the probability that the proportion of flops in a sample of 404 released films would be greater than 4%?

This is the pvlaue of Z when X = 0.04*404 = 16.16. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16.16 - 28.28}{5.13}$

$Z = -2.36$

$Z = -2.36$ has a pvalue of 0.0091

1 - 0.0091 = 0.9909

0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%

7 0

4 years ago

For a polynomial p(x), the value of p(3) is −2. Which of the following must be true about p(x)?

mr Goodwill [35]

Answer:

Step-by-step explanation:

The factor theorem states if (x - h) is a factor of a polynomial p(x) then

p(h) = 0

However, p(h) ≠ 0 then the value obtained is the remainder.

Thus

p(3) = - 2

When p(x) is divided by (x - 3) then the remainder is - 2

8 0

3 years ago

A sports club has 84 members who learned baseball, and 42 of those members also learned basketball. There are 25 students who di

HACTEHA [7]

<em>The table that shows the relative frequency of data is in the below further explanation.</em>

$\texttt{ }$

<h3>Further explanation</h3>

A set is a clearly defined collection of objects.

To declare a set can be done in various ways such as:

With words or the nature of membership

With set notation

By registering its members

With Venn diagrams

$\texttt{ }$

Multiplying set A x B is by pairing each member of set A with each member of set B.

<u>Example:</u>

Then

A x B = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}

$\texttt{ }$

Union of set A and B ( A ∪ B ) is rewriting each member A and combined with each member B.

Intersection of set A and B ( A ∩ B ) is to find the members that are both in Set A and Set B.

<u>Example:</u>

A ∪ B = {1, 2, 3, 4, 5}

A ∩ B = {3, 4}

Let us now tackle the problem!

$\texttt{ }$

<em>A sports club has 84 members who learned baseball, and 42 of those members also learned basketball.</em>

$\left[\begin{array}{ccc}&Play Baseball&Don't Play Baseball\\Play Basketball&\boxed{42}& \\Don't Play Basketball& & \end{array}\right]$

$\texttt{ }$

Number of members who learned baseball but did not learn basketball will be ( 84 - 42 ) = 42 members

$\left[\begin{array}{ccc}&Play Baseball&Don't Play Baseball\\Play Basketball&42& \\Don't Play Basketball&\boxed{42}& \end{array}\right]$

$\texttt{ }$

<em>There are 25 students who did not learn baseball but learned basketball</em>

$\left[\begin{array}{ccc}&Play Baseball&Don't Play Baseball\\Play Basketball&42&\boxed{25} \\Don't Play Basketball&42& \end{array}\right]$

$\texttt{ }$

<em>8 students did not learn either baseball or basketball</em>

$\left[\begin{array}{ccc}&Play Baseball&Don't Play Baseball\\Play Basketball&42&25 \\Don't Play Basketball&42&\boxed{8} \end{array}\right]$

$\texttt{ }$

<em>Total Number of Students = 42 + 42 + 25 + 8 = 117</em>

$\texttt{ }$

Table of relative frequency

$\left[\begin{array}{ccc}&Play Baseball&Don't Play Baseball\\Play Basketball&\boxed{\frac{42}{117}}&\boxed{\frac{25}{117}} \\Don't Play Basketball&\boxed{\frac{42}{117}}&\boxed{\frac{8}{117}} \end{array}\right]$