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3 years ago

Kasey's school starts at the time shown on the clock. what time does kasey's school start

Mathematics

1 answer:

irina1246 [14]3 years ago

4 0

Is there a picture of the clock?

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A computer company produces an extremely light laptop and they claim it is the lightest on the market weighing only 32 ounces. t

Darina [25.2K]

Solution: We are given:

$\mu=0.33, \sigma =0.7$

Let $x$ be the weight (oz) of laptop

We have to find $P(x>32)$

To find the this probability, we need to find the z score value.

The z score is given below:

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

$=\frac{32-33}{0.7}$

$=-1.43$

Now, we have to find $P(z>-1.43)$

Using the standard normal table, we have:

$P(z>-1.43)=0.9236$

0.9236 or 92.36% of laptops are overweight

4 0

3 years ago

When running a 100 meter race Bill reaches his maximum speed when he is 45 meters from the starting line and 5 seconds have elap

Mkey [24]

Bill's speed is constant after 5 seconds and 45 meters into the race.

Correct responses:

a. Bill's maximum speed is 8 m/s

b. 24 meters

c. 57 meters

<h3>Methods used to calculate speed and distance traveled</h3>

Given parameters are;

The distance of the race = 100 meter

Distance at which Bill reaches maximum speed = 45 meters

Speed Bill maintains after 5 seconds = The maximum speed

Time at which Bill is 85 meters from the starting line = 10 seconds after start

a. Required;

Bill's maximum speed in meters.

Solution:

Distance Bill runs at maximum speed, d = 85 m - 45 m = 40 m

Time at which Bill runs the 40 m at maximum speed, t = 10 s - 5 s = 5 s

$Speed = \mathbf{\dfrac{Distance}{Time}}$

Therefore;

$Bill's \ maximum \ speed = \dfrac{40 \ m}{5 \ s} = \underline{8 \ m/s}$

b. i. Required:

The distance Bill will run for 3 seconds at the maximum speed;

Solution:

Distance, s = Speed, v × Time, t

The distance traveled at maximum speed in 3 seconds is therefore;

Distance = 8 m/s × 3 s = 24 m

The distance Bill travels in 3 seconds at the maximum speed is 24 meters.

ii) The distance Bill travels at 8 seconds after start, is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 3 seconds = 24 meters

Therefore;

Bill's distance from the starting line 8 seconds after start is 45 meters + 25 meters = 69 meters

c. Bill's distance 6.5 seconds after the start of the race is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 1.5 seconds = 1.5 s × 8 m/s = 12 meters

Bill's distance from the starting line 6.5 seconds after start of the race = 45 meters + 12 meters = 57 meters

Learn more about distance, speed, time, relationship here:

https://brainly.in/question/49075584

6 0

2 years ago

Please i will give brainlist

ElenaW [278]

Answer:

the second answer

y=1/2x+3/4

5 0

3 years ago

(1/4+1/4)-6÷1/3+2 how do you get it?????

victus00 [196]

2/4-6 ÷ 1/3 + 2
|
v
2/4 - 2 + 2
|
v
2/4 -0

=2/4 -> 1/2

you get it by using PEMDAS

8 0

3 years ago

Read 2 more answers

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago