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mihalych1998 [28]
1 year ago
6

On a coordinate plane, a parallelogram has points A (negative 3, 4), B (3, 4), C (1, negative 2), and D (negative 5, negative 2)

. If a translation of (x, y) → (x + 6, y – 10) is applied to figure ABCD, what are the coordinates of D'? (–5, –2) (1, –12) (4, –15) (–9, –6)
Mathematics
2 answers:
EleoNora [17]1 year ago
5 0

Answer:

(1,-12)

Step-by-step explanation:

vladimir2022 [97]1 year ago
3 0

Answer:

(1,-12)

Step-by-step explanation:

d starts at (-5,-2) if you add 6 to -5 and minus -2 by 10 you will end up with (1,-12)

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An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "
FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

\large X_{sentences} the mean of the “sentences” group

\large S_{sentences} the standard deviation of the “sentences” group

\large X_{intentional} the mean of the “intentional” group

\large S_{intentional} the standard deviation of the “intentional” group

Then, we can calculate by using the computer

\large X_{sentences}=28.75  

\large S_{sentences}=3.53553

\large X_{intentional}=31.625

\large S_{intentional}=1.40788

\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875

The <em>standard error of the difference (of the means)</em> for a sample of size 8 is calculated with the formula

\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}

So, the standard error of the difference is

\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546

<em>In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of  95% for the difference of the means corresponding to a level of significance of 0.05 (5%). </em>

<em>If this interval contains the zero, we can say there is no significant difference. </em>

<em>Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples. </em>

We get the \large t_{0.05} which is a value of t such that the area under the Student's t distribution  outside the interval \large [-t_{0.05}, +t_{0.05}] is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

\large t_{0.05}=2.365

Now we can compute our confidence interval

\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

6 0
2 years ago
Which action will solve the equation?. m + 20 = 9. . Add 9 to each side.. Subtract 9 from each side.. Add 20 to each side.. Subt
Kruka [31]
M + 20 = 9
u need to subtract 20 from both sides because u want to isolate m

m + 20 = 9
m + 20 - 20 = 9 - 20
m = - 11
4 0
2 years ago
Read 2 more answers
The intersection of planes AMR and CMR is<br> a. MR<br> b. AC<br> c. AR<br> d. none of the above
Maurinko [17]
I believe its either a or d

3 0
2 years ago
Read 2 more answers
A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of t
erik [133]

Answer:

The speed of the jet in still air is 415 mph and the speed of the wind is 19 mph

Step-by-step explanation:

we know that

The speed is equal to divide the distance by the time

Let

x -----> the speed of the wind in miles per hour

y ----> the speed of the jet in still air in miles per hour

we know that

<em>With a tailwind</em>

y+x=\frac{868}{2}

y+x=434 ----> equation A

<em>With a headwind</em>

y-x=\frac{792}{2}

y-x=396 ----> equation B

solve the system of equations A and B by elimination

Adds equation A and equation B

y+x=434\\y-x=396\\------\\y+y=434+396\\2y=830\\y=415

<em>Find the value of x</em>

y+x=434

415+x=434

x=434-415

x=19

therefore

The speed of the jet in still air is 415 mph and the speed of the wind is 19 mph

6 0
1 year ago
Read 2 more answers
Order 93% nine tenths 0.099 0.915% and 0.092 from least to greatest
andrew11 [14]
1. 0.915% (0.00915)
2. 0.092 (9.2%)
3. 0.099 (9.9%)
4. 9/10 (.9 or 90%)
5. 93% (.93)
4 0
2 years ago
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