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3 years ago

Round 339.4749 to the nearest hundredth

Mathematics

2 answers:

ivanzaharov [21]3 years ago

6 0

The thousandth place is 4, which is less than 5 ⇒ Discard

339.4749 ≈ 339.47

Answer: 339.47

exis [7]3 years ago

4 0

Answer: 339.47 :)))))

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PLEASE HELP EMERGENCY!! IF YOU ARE NOT 100% SURE ABOUT YOUR ANSWER DO NOT ANSWER THEN!! PLEASE!!

Black_prince [1.1K]

Answer:

A -$66.65

Step-by-step explanation:

4.50x 14 = 63.00 in penalties

-63.00 + - 3.65 = -66.65

3 0

3 years ago

Read 2 more answers

Consider the probability that greater than 26 out of 124 software users will call technical support. Assume the probability that

Mekhanik [1.2K]

Answer:

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

100% probability that greater than 26 out of 124 software users will call technical support.

Step-by-step explanation:

Test if the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

It is needed that:

$np \geq 10$ and $n(1-p) \geq 10$

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

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 distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Out of 124 software users

This means that $n = 124$

Assume the probability that a given software user will call technical support is 97%.

This means that $p = 0.97$

Conditions:

$np = 124*0.97 = 120.28 \geq 10$

$n(1-p) = 124*0.03 = 3.72 < 10$

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

Consider the probability that greater than 26 out of 124 software users will call technical support.

The lowest possible probability of those is 27, so, if it is 0, since it is considerably below the mean, 100% probability of being greater. We have that:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 27) = C_{124,27}.(0.97)^{27}.(0.03)^{97} = 0$

1 - 0 = 1

100% probability that greater than 26 out of 124 software users will call technical support.

3 0

3 years ago

Dillon says that 4b and -2b are not like terms

olganol [36]

He is wrong because as long as they have the same terms/ variables behind the number it doesn’t matter about if it is positive or negative

4 0

3 years ago

How do I solve this

IgorLugansk [536]

Answer:

divide

Step-by-step explanation:

7 0

3 years ago

Which of the following statements are true?

valina [46]

The answer would be C

4 0

3 years ago