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jek_recluse [69]

2 years ago

11

Henry has collected data to find that the typing speeds for the students in a typing class has a normal distribution. What is th

e probability that a randomly selected student has a typing speed of less than 51 words per minute if the mean is 47 words per minute and the standard deviation is 4 words per minute

1 answer:

Mekhanik [1.2K]2 years ago

8 0

Answer: 0.8413

Step-by-step explanation:

Given : Henry has collected data to find that the typing speeds for the students in a typing class has a normal distribution.

Mean : $\mu=47$

Standard deviation : $\sigma= 4$

Let x be the random variable that represents the typing speeds for the students.

The z-score :-

$z=\dfrac{x-\mu}{\sigma}$

For x= 51

$z=\dfrac{51-47}{4}=1$

Using the standard normal distribution table ,the probability that a randomly selected student has a typing speed of less than 51 words per minute :-

$P(x$

Hence, the probability that a randomly selected student has a typing speed of less than 51 words per minute = 0.8413

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For an outdoor track meet to be cancelled, the temperature, t, outside must be colder than 35 degrees. Complete the following in

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2 years ago

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Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

<u>Josh:</u>

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em> $h(t)=0$ <em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

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1 year ago

After simplifying, which expressions are equivalent? Select three options.

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Answer: Second option, third option and fifth option.

Step-by-step explanation:

In order to solve this exercise it is important to remember the multiplication of signs:

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Knowing that, you can distribute the sign and add the like terms. Repeat this procedure in each expression given in the exercise.

Therefore, you get:

$a.\ (3.4a - 1.7b) + (2.5a - 3.9b)=3.4a - 1.7b + 2.5a - 3.9b=5.9a-5.6b\\\\b.\ (2.5a + 1.6b) + (3.4a + 4b)=2.5a + 1.6b + 3.4a + 4b=5.9a+5.6b\\\\c.\ (-3.9b + a) + (-1.7b + 4.9a)=-3.9b + a -1.7b + 4.9a=5.9a-5.6b\\\\d.\ -0.4b +(6b - 5.9a)=-0.4b +6b - 5.9a=6b-6.3a\\\\e.\ 5.9a - 5.6b$

You can identify that the expressions $(2.5a + 1.6b) + (3.4a + 4b)$ , $(-3.9b + a) + (-1.7b + 4.9a)$ and $5.9a - 5.6b$ are equivalent after simplifying.

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2 years ago

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bearhunter [10]

Answer:

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Step-by-step explanation:

From the question we are told that

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The selling price of a unit on the first day is s = $5

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Generally the profit of a unit on the first day is

$p_1 = 5 - 2$

$p_1 = \$3$

The profit of a unit on the second day is

$p_2 = 1 - 2$

=> $p_2 = - \$1$

Generally the probability of making profit greater than $ 1 is mathematically represented as

$P(X > 1 ) = Gamma (X ,\alpha , \theta)$

=> $P(X > 1 ) = Gamma (1 ,3 , 0.5)$

Now from the gamma distribution table we have that

$P(X > 1 ) = 0.67668$

Generally the probability of making profit less than or equal to $ 1 is mathematically represented as

$P(X \le 1 ) = 1 - P(X > 1 )$

=> $P(X \le 1 ) = 1 - 0.67668$

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and the probability of making -$1 is $P(X \le 1 ) = 0.32332$

Generally the value of profit per day is mathematically represented as

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1 year ago

A community hall is in the shape of a cuboid the hall is 40m long 15m high and 3m wide. 10 litre paint covers 25m squared costs

Darina [25.2K]

Answer:

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Step-by-step explanation:

We have been given that a community hall is in the shape of a cuboid. The hall is 40m long 15m high and 3m wide.

The paint will be required for 4 walls and ceiling.

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$\text{Area of walls and ceiling}=1410$

Therefore, the area of walls and ceiling is 1410 square meters.

Given: Cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. Therefore,

$\text{ The total painting cost}=10*(\frac{1410}{25})$

$\text{ The total painting cost}=10*56.4=564$

Therefore, the total painting cost is $564.

Tiles will be required for floor. Let us find the area of floor.

$\text{Area of floor} = 40*3\text{ square meters}$

$\text{Area of floor} =120\text{ square meters}$

Given: 1m squared floor tiles costs $3. So,

$\text{Total cost for tiles} = 3*120 = 360$

Therefore, the total tiles cost is $360.

Now let us find combined total cost of tiles and paint.

$\text{Combined total cost}= 564+360 = 924$

Therefore, the combined total cost of tiles and paint is $924.

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2 years ago