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krok68 [10]
3 years ago
8

Some help me with these two questions please!!!!!!!

Mathematics
1 answer:
Hatshy [7]3 years ago
8 0

Answer:

I believe #20. is A

Step-by-step explanation:

because if you use distributive property and multiply the 5 to the () numbers you  will get your original question.

Also, i cant see all of the options for #21.

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Complete the table and then find the function rule. Remember to put in the y=mx+b form
astraxan [27]

Answer:

1. 81        2. -20      3.  168       4.  -24

Step-by-step explanation:

I hope this is right :)

8 0
3 years ago
An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0
zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

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:

n = 105, p = 0.9

So

\mu = E(X) = np = 105*0.9 = 94.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{105*0.9*0.1} = 3.07

What is the probability that everyone who appears for the flight will get a seat

100 or less people appearing to the flight, which is the pvalue of Z when X = 100. So

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

Z = \frac{100 - 94.5}{3.07}

Z = 1.79

Z = 1.79 has a pvalue of 0.9633

96.33% probability that everyone who appears for the flight will get a seat

7 0
3 years ago
Arrange the numbers in increasing order 5.049, 5.069, 5.49, 5.09
s344n2d4d5 [400]

5.049 is the smallest

5.069

5.09

5.49 is the biggest

3 0
3 years ago
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Help now!! A triangle with side measurements 10, 24 and 26 is a right triangle.
BabaBlast [244]

Answer:

50

Step-by-step explanation:

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4 years ago
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BRAINLIEST + 30PTS<br><br> Use the form x^m • x^n = x^m+n<br><br> to solve for (x2y)(x3y4) = x^5y^5
Dimas [21]

Step-by-step explanation:

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