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Tomtit [17]
2 years ago
11

Which are true statements about a translation? Select all that apply. The image is congruent to the pre-image. The image could b

e turned upside down. The image could be moved left or right. The image could be moved up or down. The image could be a different size than the pre-image.
Mathematics
1 answer:
Novay_Z [31]2 years ago
6 0

Answer:

a} The image is congruent to the pre-image.

c} The image could be moved left or right.

d} The image could be moved up or down.

Step-by-step explanation:

i took the test

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Polygon ABCD goes through a sequence of rigid transformations to form polygon A′B′C′D′. The sequence of transformations involved
spin [16.1K]
Refer to the diagram below

The shape ABCD first is reflected on the mirror line y = x

Then it is reflected again on the y-axis [equation of the line is x = 0]

4 0
2 years ago
Read 2 more answers
A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a s
denpristay [2]

Answer:

For this case the  95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

Step-by-step explanation:

Notation

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=40-1=39

For this case the  95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

3 0
2 years ago
Round 287.9412 to the nearest tenth. Do not write extra zeros.
goldenfox [79]
Salutations!

<span>Round 287.9412 to the nearest tenth. 

Lets solve this!

To round to the nearest tenth, you need to know the tenth place--------

In the number, 287.9412, nine is in the tenth place. You need also make sure whether the number next to 9 is greater than 5 or not. 

</span><span>287.9412 =287.941=287.94=287.9=288=290

Thus, your answer is 290.

Hope I helped.
</span>
3 0
2 years ago
Read 2 more answers
Justin receives $15 and puts it into his savings account. He adds $0.25 to the account each day for a number of days, d, after t
MAVERICK [17]

Answer:

expression a

Step-by-step explanation:

The given expression is 15+0.25(d−1).

let suppose,

15 = a

0.25(d−1) = b

we get  a + b

It clearly indicates the given expression is sum of two entities, we can exclude  option b and option d.

Now we are left with option a and c, for that we have to evaluate the term b

b = 0.25(d−1) <u>that is the additional amount after d days</u>

Therefore, expression a is correct.

4 0
2 years ago
Read 2 more answers
On her first day in a hospital, Kiri receives u1 milligrams (mg) of a therapeutic drug. The amount of the drug Kiri receives inc
Valentin [98]

Answer:

a. 21 = u1 + 6·d

b. 29 = u1 + 10·d

c. d = 2, u1 = 9

Step-by-step explanation:

a. The given parameters are;

The amount of therapeutic drug Kiri receives on her first at the hospital = u1 milligrams

The amount of drug increase received by Kiri each day = d

The amount of drug Kiri received on the seventh day = 21 mg

The amount of drug she received on the eleventh day = 29 mg

Therefore, we have an arithmetic progression with the formula for the nth term given as follows;

aₙ = a₁ + (n - 1)·d

Where;

a₁ = u1

n = The number of terms

Therefore, for the 7th day, the amount of drugs she receives, which is 21 milligrams, is given as follows;

a₇ = u1 + (7 - 1)·d = u1 + 6·d = 21

The equation for the amount of drugs she receives in terms of u1 and d on the seventh day is given as follows;

21 = u1 + 6·d

b. For the eleventh day, the amount of drugs she receives, which is 29 milligrams, is given as follows;

a₁₁ = u1 + (11 - 1)·d = u1 + 10·d = 29

Therefore, the equation for the amount of drugs she receives in terms of u1 and d on the  eleventh day is given as follows;

29 = u1 + 10·d

c. Therefore, we have two equations which are given as follows;

21 = u1 + 6·d................(1)

29 = u1 + 10·d..............(2)

Subtracting equation (1) from equation (2) gives;

29 - 21 = (u1 + 10·d) - (u1 + 6·d)

8 = 4·d

d = 8/4 = 2

d = 2

From equation (1), we have;

21 = u1 + 6·d = u1 + 6×2 = u1 + 12

21 = u1 + 12

21 - 12 = u1

∴ u1 = 9

d = 2, u1 = 9.

4 0
1 year ago
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