I need help on number 11 and 18 on front of The page please Mean Median Mode and range please Thank you

Determine the sample size required to estimate the mean score on a standardized test within 4 points of the true mean with 98% c

Darina [25.2K]

Answer:

67

Step-by-step explanation:

You have to use fractions to show your work and I'm not going to do that but just know that it is the answer.

3 0

2 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

Dado el triángulo de vértices en A(-10, -13); B(-2,3) y C(2,1); Hallar la longitud de la perpendicular bajada desde el vértice B

anygoal [31]

Answer:

ghjhjhk

Step-by-step explanation:

4 0

2 years ago

A box is in the shape of an equilateral triangular prism. if the box is to be covered with paper on its lateral sides, how many

Zarrin [17]

The total surface area of the triangular prism that has a height of h and the side length of a is given below.

$\rm a(\dfrac{\sqrt3}{2} \ a + 3h)$

<h3>What is a triangular prism?</h3>

A triangular prism is a closed solid that has two parallel triangular bases connected by a rectangle surface.

A box is in the shape of an equilateral triangular prism.

If the box is to be covered with paper on its lateral sides.

Let a be the side length of the equilateral triangle and h be the height of the prism.

Then the surface area of the triangular prism will be

Surface area = 2 × area of triangle + 3 × area of the rectangle

The area of the triangle will be

$\rm Area\ of\ triangle = \dfrac{\sqrt{3}a^2}{4}$

The area of the rectangle will be

$\rm Area \ of \ rectangle = a \ h$

Then the total surface area will be

$\rm Surface\ area = 2 \times \dfrac{\sqrt3 a^2 }{4} + 3 ah\\\\\\Surface\ area = a(\dfrac{\sqrt3}{2} \ a + 3h)$

More about the triangular prism link is given below.

brainly.com/question/21308574

7 0

2 years ago

Ashley can ride her bicycle 15 miles in 2 hours. There are 60 minutes in 1 hour, and there are 1,760 yards in 1 mile. How many y

stepladder [879]

Speed is given by:
speed=distance/time
given that:
1 mile=1,760 yards
1 hour=60 min
then the speed in yards/sec will be:
distance Ashley covered is 15 miles
time=2 hours
thus;
speed=(15x1760)/(2x60)
=220 yards/ min

8 0

3 years ago

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