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3 years ago

When you have a normal distribution and you know that the area above a given value of x is .35, you also know that:

Mathematics

2 answers:

Morgarella [4.7K]3 years ago

8 0

35% of the observations have that value of x because the relative frequency of that value of x is .35.

Aloiza [94]3 years ago

5 0

Thank you for posting your question here at brainly. The answer is None of the above.

<span>35% of the observations have that value of x.
False. 0% have that value of x exactly.

35% of the observations have values less than or equal to that value of x. False . 65% of the observations have values less than or equal to that value of x.

There is a 35% chance of choosing an individual with that value of x. False . there is 0% chance of choosing an individual with exact that value of x.

The relative frequency of that value of x is .35.

False, the relative frequency is 0.</span>

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1. find the area of the figure at right. Dimensions are in inches

ioda

Answer:

292 in² (3 s.f.)

Step-by-step explanation:

The figure is made up of 2 semicircles and a rectangle.

Please see attached picture for full solution.

7 0

4 years ago

Can someone solve then explain the process of solving this?

lorasvet [3.4K]

Answer:

58°

Step-by-step explanation:

<h3>Angle sum property of triangle:</h3>

The sum of all angles of a triangle is 180°.

x + 69 + 55 + x + 78 = 180

x + x + 69 + 55 + 78 = 180

Combine like terms,

2x + 202 = 180

Subtract 202 from both sides,

2x = 180 - 202

2x = -22

Divide both sides by 2,

x = -22/2

x = -11

∠A = x + 69

= -11 + 69

$\sf \boxed{\angle A= 58^\circ}$

7 0

2 years ago

Simplify 3^8 x 3^4 divided by 3^2 x 3^8 in index form

-Dominant- [34]

3^2
Here’s the working out;

4 0

3 years ago

The expression -4.9t^2+50t + 2 represents the height, in meters, of a toy rocket t seconds after launch. The initial height of t

Kamila [148]

We have that

H(t)=-4.9t²<span>+50t + 2

we know that
</span><span>The initial height of the rocket, in meters is for t=0 sec
fot t=0
H(t)=</span>-4.9*0²+50*0 + 2------> H(t)=2 m

the answer is
The initial height of the rocket is 2 m

4 0

3 years ago

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distribute

Zina [86]

Answer:

Proportion of women having blood pressures between 88.1 and 89.4 is 3.99% or close to 4%.

Step-by-step explanation:

We are given that a recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.6.

Let X = blood pressures of adult women in the United States

So, X ~ N( $\mu=80.3,\sigma^{2} =8.6^{2}$ )

The z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean

$\sigma$ = population standard deviation

So, Probability that women have blood pressures between 88.1 and 89.4 is given by = P(88.1 < X < 89.4) = P(X < 89.4) - P(X $\leq$ 88.1)

P(X < 89.4) = P( $\frac{X-\mu}{\sigma}$ < $\frac{89.4-80.3}{8.6}$ ) = P(Z < 1.05) = 0.85314

P(X $\leq$ 88.1) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{88.1-80.3}{8.6}$ ) = P(Z $\leq$ 0.89) = 0.81327

Therefore, P(88.1 < X < 89.4) = 0.85314 - 0.81327 = 0.0399 or approx 4%

Hence, proportion of women have blood pressures between 88.1 and 89.4 is 4%.

7 0

4 years ago