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3 years ago

14

En una frabrica de eduardo hoy fabricaron 300 pantalones si los guardan en cajas donde caben 50 pantalones cuantas cajas llenaro

n

1 answer:

jek_recluse [69]3 years ago

8 0

Answer:

La respuesta a tu pregunta es sies (6)

Step-by-step explanation:

300/5 cajas = 6 cajas

300/5 boxes is equal to 6 boxes in total that are full.

Espero haber ayudado!

Hope I helped!!!

GL :)

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Draw a line representing the "run" and a line representing the "rise" of the line. State the slope of the line in simplest form.

Igoryamba

Answer:

The answer is " $\bold{\frac{9}{13}}$ "

Step-by-step explanation:

In this question, the image file is missing, which is defined in the attached file. please find it.

$\to raise= y_2-y_1$

$=-1-(-10)\\\\=-1+10\\\\=9$

$\to run= x_2-x_1$

$=9-(-4)\\\\=9+4\\\\=13$

$\to \text{formula for slope} =\frac{rise}{run} =\frac{9}{13}$

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3 years ago

What are the zeros of the function f(x) =x^2-11x+24 ?

UNO [17]

Answer:

there is none

Step-by-step explanation:

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3 years ago

Sofia makes and sells stuffed toys. She uses special fabric to cover a foam

valkas [14]

Answer:

384 square cm

Step-by-step explanation:

Fabric used will be equal to the total surface of the cubical block.

$fabric \: used \\ = 6 {side}^{2} \\ = 6 \times {8}^{2} \\ = 6 \times 64 \\ = 384 \: {cm}^{2} \\$

8 0

3 years ago

Suppose that an airline overbooks seats on their flights. In particular, it sells 300 tickets for a flight when there are only 2

vladimir1956 [14]

Using the <u>normal approximation to the binomial</u>, it is found that there is a 0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.
The binomial distribution is the probability of <u>x successes on n trials</u>, with <u>p probability</u> of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem:

15% do not show up, so 100 - 15 = 85% show up, which means that $p = 0.85$ .
300 tickets are sold, hence $n = 300$ .

The mean and the standard deviation are given by:

$\mu = np = 300(0.85) = 255$

$\sigma = \sqrt{np(1-p)} = \sqrt{300(0.85)(0.15)} = 6.185$

The probability that we will have enough seats for everyone who shows up is the probability of at most <u>270 people showing up</u>, which, using continuity correction, is $P(X \leq 270 + 0.5) = P(X \leq 270.5)$ , which is the <u>p-value of Z when X = 270.5</u>.

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

$Z = \frac{270.5 - 255}{6.185}$

$Z = 2.51$

$Z = 2.51$ has a p-value of 0.994.

0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

A similar problem is given at brainly.com/question/24261244

8 0

3 years ago

Please Help !!!!!!!!!!!!!!!!!!!!!!!!!!!!! 20% of your family likes marshmallows melted on top of the sweet potato casserole. If

Allisa [31]

Answer:

C

Step-by-step explanation:

let's say the 18 was the 100 and x was the 20% of that 100, if they equal the same just take 20% of 18 (you will get 3.6)

Therefore is C : x/18 = 20/100.

5 0

3 years ago