What is 3/5 equivalent too ?

Consider a normal distribution with mean 23 and standard deviation 7. What is the probability a value selected at random from th

Trava [24]

Answer:

0.5 = 50% probability a value selected at random from this distribution is greater than 23

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 23, \sigma = 7$

What is the probability a value selected at random from this distribution is greater than 23?

This is 1 subtracted by the pvalue of Z when X = 23. So

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

$Z = \frac{23 - 23}{7}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.5 = 50% probability a value selected at random from this distribution is greater than 23

5 0

3 years ago

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A ball dropped from the top of a building can be modeled by the function f(t)=-16^2+36, where t represents time in seconds

Viefleur [7K]

Answer:

The ball is dropped from a height of 36 feet

The bee and the ball will collide after approximately 1.3 seconds

The bee and the ball will collide at approximately 8 feet above the ground

The ball hits the ground after 1.5 seconds

Step-by-step explanation:

The complete question is

A ball dropped from the top of the building can be modeled by the function f(t)=-16t^2 + 36 , where t represents time in seconds after the ball was dropped. A bee's flight can be modeled by the function, g(t)=3t+4, where t represents time in seconds after the bee starts the flight.

The graph represents the situation.

select all that apply

1) The bee launches into flight from the ground.

2) The ball is dropped from a height of 36 feet.

3) The bee and the ball will collide after approximately 1.3 seconds.

4) The bee and the ball will collide after approximately 8 seconds.

5) The bee and the ball will collide at approximately 8 feet above the ground.

6) The bee and the ball will not collide.

7) The ball hits the ground after 1.5 seconds

see the attached figure to better understand the problem

Verify all the options

case 1) The bee launches into flight from the ground.

The statement is false

Because

we have

$g(t)=3t+4$

For t=0

$g(t)=3(0)+4=4\ ft$

That means ----> The bee launches into flight from a height 4 feet above the ground

case 2) The ball is dropped from a height of 36 feet

The statement is true

Because

For t=0

$f(t)=-16(0)^2+36=36\ ft$

case 3) The bee and the ball will collide after approximately 1.3 seconds

The statement is true

Because

Equate f(t) and g(t)

$-16t^2+36=3t+4$

$-16t^2-3t+32=0$

solve the quadratic equation by graphing

The solution is t=1.324 sec

see the attached figure N 2

case 4) The bee and the ball will collide after approximately 8 seconds

The statement is false

Because, the bee and the ball will collide after approximately 1.3 seconds (see case 3)

case 5) The bee and the ball will collide at approximately 8 feet above the ground

The statement is true

Because

For t=1.324 sec

substitute the value of t in f(t) or g(t)

$g(t)=3(1.324)+4=7.97\ ft$

case 6) The bee and the ball will not collide

The statement is false

see case 3) and case 5)

case 7) The ball hits the ground after 1.5 seconds

The statement is true

Because

we know that

The ball hit the ground when the value of f(t) is equal to zero

so

For f(t)=0

$-16t^2+36=0$

$t^2=\frac{36}{16}$

$t=\frac{6}{4}=1.5\ sec$

4 0

2 years ago

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Find the midpoint of the segment with the following endpoints. (8,5) and (0, -1)

Karo-lina-s [1.5K]

Answer:

(4,2)

Step-by-step explanation:

$Midpoint =\left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)\\\\\left(x_1,\:y_1\right)=\left(8,\:5\right),\:\\\left(x_2,\:y_2\right)=\left(0,\:-1\right)\\\\=\left(\frac{0+8}{2},\:\frac{-1+5}{2}\right)\\\\= (\frac{8}{2} , \frac{4}{2} )\\\\Simplify\\=\left(4,\:2\right)$