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shepuryov [24]
3 years ago
6

Solve quadratic equations by factoring x cubed +3x squared - 4x= 0

Mathematics
2 answers:
Papessa [141]3 years ago
8 0
<span>x cubed +3x squared - 4x= 0
= x^3 + 3x^2 -4x = 0
=x(x^2 +3x -4) = 0
= x (x+4)(x-1) = 0

x = 0
x+ 4 = 0 then x = -4
x - 1 = 0 then x = 1
answer: x = 0, x = -4 and x = 1</span>
Mumz [18]3 years ago
6 0

<span>x(x^2-3x-4)=0 </span>
<span>factor </span>
<span>x(x-4)(x+1)=0 </span>

<span>x=0 </span>
<span>x=4 </span>
<span>x=-1</span>
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Assume that females have pulse rates that are normally distributed with a mean of beats per minute and a standard deviation of b
snow_tiger [21]

Answer:

a) This is the p-value of Z when X = A subtracted by the p-value of Z when X = B.

b) P-value of Z when X = 76 subtracted by the p-value of X = 68.

c) Because the underlying distribution(pulse rates of females) is normal.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal 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.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

Mean \mu, standard deviation \sigma.

standard deviation of beats per minute.

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is between A beats per minute and B beats per minute.

This is the p-value of Z when X = A subtracted by the p-value of Z when X = B.

b. If 4 adult females are selected, find the probability that they have pulse rates with a mean between 68 beats per minute and 76 beats per minute.

Sample of 4 means that we have n = 4, s = \frac{\sigma}{\sqrt{4}} = 0.5\sigma

The formula for the z-score is:

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

Z = \frac{X - \mu}{0.5\sigma}

Z = 2\frac{X - \mu}{\sigma}

This probability is the p-value of Z when X = 76 subtracted by the p-value of X = 68.

c. Why can the normal distribution be used in heartbeat even the sample side does not exceed 30?

Because the underlying distribution(pulse rates of females) is normal.

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*
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okay i have question is this more than oonequestion Step-by-step explanation:

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