Evaluate cos 90°. Round your answer to the nearest hundredth if necessary

For a normal distribution with μ=500 and σ=100, what is the minimum score necessary to be in the top 60% of the distribution?

STatiana [176]

Answer:

475.

Step-by-step explanation:

We have been given that for a normal distribution with μ=500 and σ=100. We are asked to find the minimum score that is necessary to be in the top 60% of the distribution.

We will use z-score formula and normal distribution table to solve our given problem.

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

Top 60% means greater than 40%.

Let us find z-score corresponding to normal score to 40% or 0.40.

Using normal distribution table, we got a z-score of $-0.25$ .

Upon substituting our given values in z-score formula, we will get:

$-0.25=\frac{x-500}{100}$

$-0.25*100=\frac{x-500}{100}*100$

$-25=x-500$

$-25+500=x-500+500$

$x=475$

Therefore, the minimum score necessary to be in the top 60% of the distribution is 475.

3 0

2 years ago

A fitness center runs a contest for customers who enroll in its annual membership plan. The customers are allowed to draw a coup

krek1111 [17]

1. 20/35(30)=17.142

2. 5/35(60)=17.142

3. 10/35(60)=17.142

6 0

3 years ago

Read 2 more answers

A factory produce bicycles at a rate of 110+0.5t^2-0.9t bicycles per week (t in weeks)

klasskru [66]

Answer:

221

Step-by-step explanation:

Day 8 is the first day of the second week.

Day 21 is the last day of week 3.

We need to know the n umber of bicycles made from t = 1 to t = 3

The function is b(t) = 110 + 0.5t^2 - 0.9t, where t is in weeks.

We need to integrate the function with the limits of 1 to 3.

$\int_{1}^{3} (110 + 0.5t^2 - 0.9t) dt$

$\int_{1}^{3} (110 + \dfrac{t^2}{2} - \dfrac{9t}{10}) dt$

$= 110t + \dfrac{t^3}{6} - \dfrac{9t^2}{20} \Biggr|_{1}^{3}$

$= 330.45 - 109.72$

$= 220.7333$

Answer: 221

3 0

3 years ago

A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of t

Makovka662 [10]

Answer:

132.233 ft2

Step-by-step explanation:

Let's call the width of the rectangle 'w' and the length 'x'. So the area of the semicircle is:

$A_1 = \pi*radius^2/2$

$A_1 = \pi*(w/2)^2/2$

$A_1 = \pi/8*w^2$

And the area of the rectangle is:

$A_2 = w*x$

If the perimeter of the window is 41 feet, we have:

$Perimeter = length + 2*width + \pi*radius$

$41 = x + 2*w + \pi*w/2$

$x = 41 - w(2 + \pi/2)$

Now, the equation for the total area of the window is:

$A = A_1 + A_2 = \pi/8*w^2 + w*x$

$A = \pi/8*w^2 + w*(41 - w(2 + \pi/2))$

$A = (\pi/8-2 - \pi/2)*w^2 + 41w = -3.1781w^2 + 41w$

To find the maximum area, we can find the x-coordinate of the vertex of the quadratic equation:

$x\_vertex = -b / 2a = -41 / (-3.1781*2) = 6.45$

So the width that gives us the maximum area of the window is 6.45 feet, and the area will be:

$A = -3.1781w^2 + 41w = -3.1781*(6.45)^2 + 41*6.45 = 132.233\ ft^2$

3 0

3 years ago

Consider a single spin of the spinner. Which events are mutually exclusive? Choose all that apply. a)landing on a shaded sector

denis23 [38]

Answer: the correct options are b and d.

Step-by-step explanation:

Let us first define what mutually exclusive events are. If two events are mutually exclusive, it means that they cannot happen at the same time. For example, getting a head and a tail at the same time are mutually exclusive.

Considering a single spin of the spinner, the events that are mutually exclusive are

b)landing on a shaded sector and landing on a 3. This is because 3 is unshaded. You can either land on 3 or an unshaded sector at a time

d)landing on an unshaded sector and landing on a number less than 2. This is because the numbers in the unshaded sectors are greater than 2.

3 0

3 years ago