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3 years ago

Can somebody help plzzzz anyone

Mathematics

2 answers:

aliya0001 [1]3 years ago

4 0

Answer:

The answer is D.

Step-by-step explanation:

Look at the amount of zeros went over to the left so negative 3

Katen [24]3 years ago

4 0

Answer- D because it’s takes 3 integers in the negative way to get to the 3 in .0013
So it’ll be 10^-3

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An open box is made from an 8 by ten-inch rectangular piece of cardboard by cutting squares from each corner and folding up the

natka813 [3]

Please find the attachment.

Let x represent the side length of the squares.

We have been given that an open box is made from an 8 by ten-inch rectangular piece of cardboard by cutting squares from each corner and folding up the sides. We are asked to find the volume of the box.

The side of box will be $8-x-x=8-2x$ and $10-x-x=10-2x$ .

The height of the box will be $x$ .

The volume of box will be area of base times height.

$\text{Volume of box}=(8-2x)(10-2x)\cdot x$

Now we will use FOIL to simplify our expression.

$\text{Volume of box}=(80-16x-20x+4x^2)\cdot x$

$\text{Volume of box}=(80-36x+4x^2)\cdot x$

Now we will distribute x.

$\text{Volume of box}=80x-36x^2+4x^3$

$V(x)=80x-36x^2+4x^3$

Therefore, the volume of the box would be $V(x)=80x-36x^2+4x^3$ .

6 0

3 years ago

kicyunya [14]

Answer:

8.69% probability that at least 285 rooms will be occupied.

Step-by-step explanation:

For each booked hotel room, there are only two possible outcomes. Either there is a cancelation, or there is not. So we use concepts of the binomial probability distribution to solve this question.

However, we are working with a big sample. So i am going to aproximate this binomial distribution to the normal.

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:

A popular resort hotel has 300 rooms and is usually fully booked. This means that $n = 300$

About 7% of the time a reservation is canceled before the 6:00 p.m. deadline with no pen-alty. What is the probability that at least 285 rooms will be occupied?

Here a success is a reservation not being canceled. There is a 7% probability that a reservation is canceled, and a 100 - 7 = 93% probability that a reservation is not canceled, that is, a room is occupied. So we use $p = 0.93$