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Serggg [28]
3 years ago
14

a rectangle has an area of 1/6 squar centimeters and a length of 1.5 centimeters what is the width and the perimeter

Mathematics
1 answer:
Alex73 [517]3 years ago
3 0
<span>

(1/6)/ 1.5 = 
(1/6)/(15/10)=
1/6*10/15=10/90=1/9 width</span>
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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, <em>t</em> = 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,<em> s</em> = Speed, <em>v</em> × Time, <em>t</em>

<em />

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 = <u>69 meters</u>

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 = <u>57 meters</u>

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

https://brainly.in/question/49075584

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Dennis arrived at Sullivan's at 10:15am the store opens at 11:00 how long will he have to wait
steposvetlana [31]
He will have to wait 45 minutes
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4 years ago
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Find the slope of the line through the pair of points. A(2, 3), P(2, 9)
AVprozaik [17]
A(x_1;\ y_1);\ B(x_2;\ y_2)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\-------------------\\\\A(2;\ 3)\to x_1=2\ and\ y_1=3\\B(2;9)\to x_2=2\ and\ y_2=9\\\\m=\dfrac{9-3}{2-2}=\dfrac{7}{0}-divided\ by\ 0\\\\Answer:\ The\ slope\ unde fined. It's\ a\ vertical\ line\ x=2.
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3 years ago
The scale on a map indicates that 1 inch corresponds to an actual distance of 95 miles. two cities are 4.5 inches apart on the m
arlik [135]
The two cities are actually 427.5 miles apart.
3 0
3 years ago
a random sample of 510 high school students has a normal distribution. The sample mean average ACT exam score was 21 with a 3.2
34kurt

Answer:

CI = 21 ± 0.365

Step-by-step explanation:

The confidence interval is:

CI = x ± SE * CV

where x is the sample mean, SE is the standard error, and CV is the critical value (either t score or z score).

Here, x = 21.

The standard error for a sample mean is:

SE = σ / √n

SE = 3.2 / √510

SE = 0.142

The critical value is looked up in a table or found with a calculator.  But first, we must find the alpha level and the critical probability.

α = 1 - 0.99 = 0.01

p* = 1 - (α/2) = 1 - (0.01/2) = 0.995

Using a calculator or a z-score table:

P(x<z) = 0.995

z = 2.576

Therefore:

CI = 21 ± 0.142 × 2.576

CI = 21 ± 0.365

Round as needed.

8 0
4 years ago
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