Find the distance between (–3, –3) and (6, 12). Round the answer to the nearest hundredth, if necessary.

2 answers:

7 0

$A=(x_A;y_A)\ \ \ and\ \ \ B=(x_B;y_B)\\\\ \Rightarrow\ \ \ |AB|= \sqrt{(x_A-x_B)^2+(y_A-y_B)^2} \\----------------------\\\\A=(-3;-3);\ \ \ B=(6;12)\\\\|AB|= \sqrt{(-3-6)^2+(-3-12)^2} =\\\\.\ \ \ \ \ \ = \sqrt{81+225} = \sqrt{306} =3 \sqrt{34} \approx17.49$

marin [14]3 years ago

5 0

$(-3, -3), \ \ (6, 12)\\ \\distance \ formula : \\ \\d =\sqrt{(x_{2}-x_{1})^2 +(y_{2}-y_{1})^2} \\ \\d =\sqrt{(6-(-3))^2 +(12-(-3) )^2} =\sqrt{(6+3)^2 +(12+3 )^2} =\sqrt{9^2+15^2}=\\ \\=\sqrt{81+225}=\sqrt{306}=\sqrt{9\cdot 34}=3\sqrt{34}=3\cdot 5.83= 17.49$

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A statistics student wants to compare his final exam score to his friend's final exam score from last year; however, the two exa

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Answer:

$z= \frac{85-70}{10}=1.5$

$z= \frac{45-35}{5}=2$

So then the correct answer for this case is:

B) Our student, Z= 1.50; his friend, Z=2.00.

Step-by-step explanation:

Assuming this complete question:

A statistics student wants to compare his final exam score to his friend's final exam score from last year; however, the two exams were scored on different scales. Remembering what he learned about the advantages of Z scores, he asks his friend for the mean and standard deviation of her class on the exam, as well as her final exam score. Here is the information:

Our student: Final exam score = 85; Class: M = 70; SD = 10.

His friend: Final exam score = 45; Class: M = 35; SD = 5.

The Z score for the student and his friend are:

A) Our student, Z= -1.07; his friend, Z= -1.14.

B) Our student, Z= 1.50; his friend, Z=2.00.

C) Our student, Z= 1.07; his friend, Z= -1.14.

D) Our student, Z= 1.07; his friend, Z= 1.50

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution

Let X the random variable that represent the scores for our student, and for this case we know that:

$E(X)= \mu =70, SD_X=\sigma=10$