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r-ruslan [8.4K]
3 years ago
11

Use words to write a comparison statement for the problem above.

Mathematics
2 answers:
Drupady [299]3 years ago
7 0

The two number which we have to compare is

1.→ 41,253

2.→ 35,214

The first number as well as Second number is a 5 digit number.

The Unit Digit of first number is 4 and unit digit of second number is 3.So, Unit place of first five digit number is greater than unit place of Second five digit number.

So, We can say that

  → 41,253 > 35,214

Comparison Statement

There are two five digit Number made from digits 1,2,3,4,and 5.The two number which we have to compare using <,= and >, is, 41,253 and 35,214.

pantera1 [17]3 years ago
6 0

Looking at the question we can see that it says:

41253>35214

We should know what these symbols mean:

stands for something less than the other, for example:

We can say that 2 is less than 5 so in a symbolic form it can be written as,

2

Similarly,

> stands for something greater than the other, for example:

We can say that 7 is greater than 5, in a symbolic form it can be written as,

7>5

So here, 42153>35214 means that, 42153 is greater than 35214.

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In which number does the 6 have a value that is one tenth the value of the 6 in 34, 761
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- у<br> 6x - 4у - 2<br><br> What is the next step?
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3 years ago
Suppose that 35 people are divided in a random manner into two teams in such a way that one team contains 10 people and the othe
Effectus [21]

Answer:

0.5798 or 57.98%

Step-by-step explanation:

The total number of ways to form the two teams is the combination of choosing 10 people out of 35 (₃₅C₁₀). The number of possibilities that A and B are both on the 10-people team is given by the combination of choosing 8 people (since two are fixed) out of 33 (₃₃C₈).The number of possibilities that A and B are both on the 25-people team is given by the combination of choosing 10 people out of 33 (₃₃C₈).

Therefore, the probability that two particular people A and B will be on the same team is:

P = \frac{_{33}C_8+_{33}C_{10}}{_{35}C_{10}} \\\\P=\frac{\frac{33!}{(33-8)!8!}+\frac{33!}{(33-10)!10!} }{\frac{35!}{(35-10)!10!}} \\\\P=\frac{69}{119} = 0.5798

The probability is 0.5798 or 57.98%.

8 0
3 years ago
P(X&lt; ) 1-P(X&gt; ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball
Mashutka [201]

Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

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 = 30, p = 0.626

So

\mu = E(X) = np = 30*0.626 = 18.78

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65

What is the probability that no more than 16 of them are strikes?

Using continuity correction, this is P(X \leq 16 + 0.5) = P(X \leq 16.5), which is the pvalue of Z when X = 16.5. So

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

Z = \frac{16.5 - 18.78}{2.65}

Z = -0.86

Z = -0.86 has a pvalue of 0.1949

19.49% probability that no more than 16 of them are strikes

7 0
3 years ago
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