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3 years ago

A symbol that represents and unknown quantity is called ??

Mathematics

2 answers:

Natasha_Volkova [10]3 years ago

7 0

the answer is a variable

Tom [10]3 years ago

4 0

Answer:

Variables

Step-by-step explanation:

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Help pls 6th grade math

vladimir1956 [14]

Answer:

x> or equal to 35. 36 is a solution.

Step-by-step explanation:

3 0

2 years ago

Y=x+2 and then you have x=-3 You also have to use transitive property

tresset_1 [31]

If y = x+2 and x = -3 then we can say that y = -3+2 after replacing the 'x' with -3

From here, we just add -3 and 2 to get -1 which is why y = -3+2 turns into y = -1

So together x = -3 and y = -1

They pair up to form (x,y) = (-3,-1)

8 0

3 years ago

An array of 2 rows of 7 dots has a total of 14 dots. it is called a 2 by __ array

max2010maxim [7]

It is called a 2 by 7 array.

3 0

3 years ago

Read 2 more answers

I need help with all three

a_sh-v [17]

The cactus one I think is 0.875

8 0

3 years ago

Read 2 more answers

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 normal approximation to the binomial, 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 x successes on n trials, with p probability 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 270 people showing up, which, using continuity correction, is $P(X \leq 270 + 0.5) = P(X \leq 270.5)$ , which is the p-value of Z when X = 270.5.

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

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

$Z = 2.51$

$Z = 2.51$ 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