How do I solve this???

Round this number to the nearest million<br><br> 135,458,267

Svet_ta [14]

1,000,000 is million
to round look at hundred thousand place or 100,000
see number before million (after comma)

if it is greater than or equal to 5, add 1 to current value of million and say all after is 0
if less than 5 then leave million value as is and make after 0

135,458,267
hundred thousand value is 4
4<5
135,000,000

4 0

3 years ago

A chemical factory has 2 cylindrical chemical tanks, one containing Chemical X and the other containing Chemical Y. The tank con

Mashutka [201]

The tank with Chemical X "takes up" a space of 25ft³. Ordinarily we think of something "taking up" space as being area or surface area; however, area is a square measurement, and this is cubic; this must be volume. The volume of the tank with Chemical X is 1.5 times the volume of the tank containing Chemical Y; setting this up in an equation we would have
25 = 1.5<em>V</em>
We would divide both sides by 1.5 to get the volume of the tank containing Chemical Y:
$\frac{25}{1.5}=\frac{1.5V}{1.5} \\16 \frac{2}{3}=V$
To find the volume of a cylinder, we find the base area and multiply by the height. We know the volume and we know the base area, so our equation to find the height of the tank containing Chemical Y would look like:
$16 \frac{2}{3}=3.2h \\ 16 \frac{2}{3}=3 \frac {2}{10}h$
We would now divide both sides by 3 2/10:
$\frac{16 \frac{2}{3}}{3 \frac{2}{10}}= \frac{3 \frac{2}{10}h}{3 \frac{2}{10}}$
This is the same as:
$\frac{\frac{50}{3}}{\frac{32}{10}}=h \\ \\ \frac{50}{3}* \frac{10}{32}=h \\ \\ \frac{500}{96}$
So the height of the tank containing Chemical Y is 500/96 = 5 5/24 feet.

8 0

3 years ago

Calculate the mean and standard deviation for the average yards rushed by

Amiraneli [1.4K]

Question:

A sample of cans of peaches was taken from a warehouse, and the content of each can mearsed for weight. the sample means was 486g with stand deviation 6g. state the weight percentage of cans with weight:

Draw normal curve to help - split into 8 section ( i can't draw it here)

a) 34.13% of cans will be between 480g and 486g.

b) 13.59 + 2.15 + 0.13= 15.87% of cans greater than 492g.

Look at the Stand Dev Graph where the give you the number

Step-by-step explanation:

Let me give you a different question and i will answer it, which you help you answer your question.

8 0

2 years ago

In a beach town, 13% of the residents own boats. A random sample of 100 residents was selected. What is the probability that les

mariarad [96]

Answer:

Approximately $0.038$ (or equivalently $3.8\%$ ,) assuming that whether each resident owns boats is independent from one another.

Step-by-step explanation:

Assume that whether each resident of this town owns boats is independent from one another. It would be possible to model whether each of the $n = 100$ selected residents owns boats as a Bernoulli random variable: for $k = \lbrace 1,\, \dots,\, 100\rbrace$ , $X_k \sim \text{Bernoulli}(\underbrace{0.13}_{p})$ .

$X_k = 0$ means that the $k$ th resident in this sample does not own boats. On the other hand, $X_k = 1$ means that this resident owns boats. Therefore, the sum $(X_1 + \cdots + X_{100})$ would represent the number of residents in this sample that own boats.

Each of these $100$ random variables are all independent from one another. The mean of each $X_k$ would be $\mu = 0.13$ , whereas the variance of each $X_k\!$ would be $\sigma = p\, (1 - p) = 0.13 \times (1 - 0.13) = 0.1131$ .

The sample size of $100$ is a rather large number. Besides, all these samples share the same probability distribution. Apply the Central Limit Theorem. By this theorem, the sum $(X_1 + \cdots + X_{100})$ would approximately follow a normal distribution with:

mean $n\, \mu = 100 \times 0.13 = 13$ , and
variance $\sigma\, \sqrt{n} = p\, (1 - p)\, \sqrt{n} = 0.1131 \times 10 = 1.131$ .

$11\%$ of that sample of $100$ residents would correspond to $11\% \times 100 = 11$ residents. Calculate the $z$ -score corresponding to a sum of $11$ :

$\begin{aligned}z &= \frac{11 - 13}{1.131} \approx -1.77 \end{aligned}$ .

The question is (equivalently) asking for $P( (X_1 + \cdots + X_{100}) < 11)$ . That is equal to $P(Z < -1.77)$ . However, some $z$ -tables list only probabilities like $P(Z > z)$ . Hence, convert $P(Z < -1.77)\!$ to that form: