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3 years ago

Geometry PEOPLE HELP

Mathematics

1 answer:

White raven [17]3 years ago

6 0

Answer: second option.

Step-by-step explanation:

Given the transformation $T:(x,y)$ → $(x-5,y+3)$

You must substitute the x-coordinate of the point A (which is $x=2$ ) and the y-coordinate of the point A (which is $y=-1$ ) into $(x-5,y+3)$ to find the x-coordinate and the y-coordinate of the image of the point A.

Therefore, you get that the image of A(2,-1) is the following:

$(x-5,y+3)=(2-5,-1+3)=(-3,2)$

You can observe that this matches with the second option.

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The variable z is inversely proportional to x. When x is 6, z has the value 0.5. What is the value of z when x= 14?

Lilit [14]

Z is inversely proportional to X is the same as
z is directly proportional to 1/X
which can be written like this:-
y=k/X
________
0.5=k/6
0.5*6=k/6*6
k=3 ;
________
z = 3/14
Z = 0.2

7 0

3 years ago

Solve for − ∣ x − 2 ∣ + 4 ≤ 0

MakcuM [25]

Answer:

x ≥ 6 or x ≤ -2

Step-by-step explanation:

- I x - 2 I + 4 ≤ 0

- I x - 2 I ≤ -4

I x - 2 I ≥ 4

x - 2 ≥ 4 or x - 2 ≤ -4

x ≥ 6 or x ≤ -2

5 0

3 years ago

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95