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3 years ago

PLZ HELP ASAP NEED HELP

Mathematics

1 answer:

SashulF [63]3 years ago

4 0

Answer: $x_1=2\\\\x_2=-\frac{1}{2}$

Step-by-step explanation:

To find the zeros of the quadratic equation given in the exercise, you need to apply the proccedure shown below:

- You must rewrite it as following:

$6x^2-9x-6=0$

-You can apply the Quadratic formula, which is shown below:

$x=\frac{-b\±\sqrt{b^2-4c}}{2a}$

In this case:

$a=6\\b=-9\\c=-6$

- Therefore, when you substitute values, you obtain:

$x=\frac{-(-9)\±\sqrt{(-9)^2-4(6)(-6)}}{2(6)}$

$x_1=2\\\\x_2=-\frac{1}{2}$

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Find slope and y-intercept<br> 5x+3y=-6

katen-ka-za [31]

Answer:

$m=-\frac53; q=-2$

Step-by-step explanation:

Let's bring the line in the classical $y=mx+q$ form

$3y= -5x-6\\y=-\frac53 x -2$

At this point we have the two values $-\frac53$ for the slope and $2$ for the intercept

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3 years ago

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

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Step-by-step explanation:

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