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yulyashka [42]
3 years ago
15

All sets that contain the number 5 ?

Mathematics
1 answer:
vivado [14]3 years ago
5 0
Real numbers, rational numbers, Integers, Whole numbers, and Natural numbers. 
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Can someone see if this is right
iren [92.7K]
Correct, the answer is B.
5 0
3 years ago
Read 2 more answers
If x=5 then what is x+1278436-888+946286-x+7+x?
Anvisha [2.4K]
Well if x=5 then you just plug in so 5+<span>1278436-888+946286-5+7+5= 2223846
Just go from left to right when answering this problem.</span>
4 0
3 years ago
Please Help!! What is the next step in the proof? Choose the most logical approach.
marta [7]

Answer:

statement D is correct.

<em>Statement: ∠7 ≅ ∠6 and ∠8 ≅ ∠5</em>

<em>Reason: Vertical Angles Theorem</em>

Step-by-step explanation:

Given that

3 ≅ ∠7 and ∠4 ≅ ∠8

from statement 2 because they are corresponding angles.

∠8 ≅ ∠5

because its vertical angles

The vertical angles theorem is about angles that are opposite each other.

So,

∠4 ≅ ∠8 and ∠8 ≅ ∠5

which means

<h3> ∠4 ≅ ∠5 </h3>

Hence,

∠3 = ∠7

∠4 = ∠5

they are known as interior alternative angles.

7 0
3 years ago
The area of a rectangle is represented with the expression (18x-12) inches squared. The width of the rectangle is 6 inches. Writ
Rudik [331]
Are equation for a rectangle is area = (width x length)

(18x - 12) = 6 x length
length = (18x - 12) / 6
length = 3x - 2
5 0
2 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
2 years ago
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