6.SP.2 Understand that a set of data collected to answer a statistical question has a distribution which can be described by its

mina [271]

Just multiply all them together for volume, 468 ft ^3

8 0

3 years ago

W trójkącie ABC połączono wierzchołek C z punktem D który jest środkiem boku AB . Wyznaczono równierz środek odcinka CD punkt E

qwelly [4]

Translate into English please

8 0

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.

5 0

3 years ago

1) A clerk have y dollars, all in $10 bills. What expression repression the number of $10 bills the clerk has?

valina [46]

1) y:10;
2) 4p+6q;
3) x +( x+ 1) = 59 => 2x = 58 => x = 29 => 29 and 30;

4 0

3 years ago

Read 2 more answers

Waves with an amplitude of 2 feet pass a dock every 30 seconds. Write an equation for a cosine function to model the height of a

kolbaska11 [484]

Answer:

The cosine function to model the height of a water particle above and below the mean water line is h = 2·cos((π/30)·t)

Step-by-step explanation:

The cosine function equation is given as follows h = d + a·cos(b(x - c))

Where:

$\left | a \right |$ = Amplitude

2·π/b = The period

c = The phase shift

d = The vertical shift

h = Height of the function

x = The time duration of motion of the wave, t

The given data are;

The amplitude $\left | a \right |$ = 2 feet

Time for the wave to pass the dock

The number of times the wave passes a point in each cycle = 2 times

Therefore;

The time for each complete cycle = 2 × 30 seconds = 60 seconds

The time for each complete cycle = Period = 2·π/b = 60

b = π/30 =

Taking the phase shift as zero, (moving wave) and the vertical shift as zero (movement about the mean water line), we have

h = 0 + 2·cos(π/30(t - 0)) = 2·cos((π/30)·t)

The cosine function is h = 2·cos((π/30)·t).

4 0

3 years ago