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mote1985 [20]
3 years ago
14

Independent Practice

Mathematics
1 answer:
gladu [14]3 years ago
4 0

Answer:

D.

d = 3t; 90 mi

Step-by-step explanation:

Bicycling A bicyclist traveled at a constant speed during a timed practice period. Write a proportion to find the distance the cyclist traveled in 30 min.

elapsed time    distance

10 min               3 mi

25 min               7.5 mi

Graph it. :)

Start at (0,0), then at 10 min. on the x-axis, go 3 spaces up on the y-axis and plot a point. Do the same for 25 min. and 7.5 miles. Draw a line through all of those points, and see where the line hits on the y-axis when the x-axis is at 30 min. :)

If two things vary directly it means their ratio is a constant.  Therefore from the formula rt=d. r=d/t,  r=3/10 or 7.5/25  which reduces to 3/10.  This is the constant speed,the lable is mi/min.  Now use  (d/30)=(3/10)  and solve for d.  Cross multiply and 10d=90, or d=9 mi

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klasskru [66]
-4(3)=-12
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3 0
3 years ago
Read 2 more answers
PLEASE HELP PLEASE 9TH GRADE MATH
wel

Answer:

A

Step-by-step explanation:

I believe it is A because for perpendicular lines the slope must be the reciprocal and have the opposite sign. The y-intercept 2 matches the coordinates (-2, 2). The second number, 2, is the y coordinate.

4 0
3 years ago
A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed
AURORKA [14]

Answer:

A 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Step-by-step explanation:

We are given that Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured.

For the 25-mil film, the sample data result is: Mean Standard deviation 1.15 0.11 and For the 20-mil film the data yield: Mean Standard deviation 1.06 0.09.

Firstly, the pivotal quantity for finding the confidence interval for the difference in population mean is given by;

                     P.Q.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean speed for the 25-mil film = 1.15

\bar X_1 = sample mean speed for the 20-mil film = 1.06

s_1 = sample standard deviation for the 25-mil film = 0.11

s_2 = sample standard deviation for the 20-mil film = 0.09

n_1 = sample of 25-mil film = 8

n_2 = sample of 20-mil film = 8

\mu_1 = population mean speed for the 25-mil film

\mu_2 = population mean speed for the 20-mil film

Also,  s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} } = \sqrt{\frac{(8-1)\times 0.11^{2}+ (8-1)\times 0.09^{2}}{8+8-2} } = 0.1005

<em>Here for constructing a 98% confidence interval we have used a Two-sample t-test statistics because we don't know about population standard deviations.</em>

<u>So, 98% confidence interval for the difference in population means, (</u>\mu_1-\mu_2<u>) is;</u>

P(-2.624 < t_1_4 < 2.624) = 0.98  {As the critical value of t at 14 degrees of

                                             freedom are -2.624 & 2.624 with P = 1%}  

P(-2.624 < \frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } < 2.624) = 0.98

P( -2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } < 2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } <  ) = 0.98

P( (\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } ) = 0.98

<u>98% confidence interval for</u> (\mu_1-\mu_2) = [ (\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } , (\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } } ]

= [ (1.15-1.06)-2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } } , (1.15-1.06)+2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } } ]

 = [-0.042, 0.222]

Therefore, a 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Since the above interval contains 0; this means that decreasing the thickness of the film doesn't increase the speed of the film.

7 0
3 years ago
The width is 7 and the diagonal is 25 what is the height
Korvikt [17]
The height is 24.

We can use Pythagorean’s Theorem: a^2+b^2=c^2

substituting values we get:

a^2+7^2=25^2

Solving we get:
a^2+49=625
a^2=576
a=24.

So the height is 24.

Hope this helps!
8 0
3 years ago
A model giving the area A, in square meters, of a path is A = x^2 - 18x +80. Which equivalent expression shows the length and th
levacccp [35]

Answer:

B

Step-by-step explanation:

4 0
2 years ago
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