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3 years ago

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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3 years ago

Use the regression line for the data in the scatterplot to answer the question. On average, how many more goals will a player sc

kari74 [83]

Answer:

A. 1/2 of a goal

Step-by-step explanation:

Hello!

Given the options:

A. 1/2 of a goal

B. 3/4of a goal

C. 1 goal

D. 4/3 goals

In a linear regression analysis the interpretation of the slope is:

> The modification of the average of Y when X increases one unit.

Given the variables

Y: Goals scored in the tournament.

X: Hours of training.

To calculate the slope, you have to use the y-intercept and choose two pairs of (Y;X) values that correspond to a value on the line, for example (1;1.5) and (2;1.5).

You can calculate the slope as b= ΔY/ΔX

$b= \frac{Y_2-Y_1}{X_2-X_1}= \frac{1.5-1}{1-0} = \frac{0.5}{1}= \frac{1}{2}$

1st value (X₁;Y₁)= (0;1)

2nd value (X₂Y₂)= (1;1.5)

So the average goals a player scores for every additional hour of practice is 1/2.

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3 years ago

Read 2 more answers

The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

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3 years ago

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Bond [772]

Energy transfer is your answer <span />

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3 years ago

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Simplify completely:<br> 4V10 – V6 +3710<br> Show all work for full credit.

schepotkina [342]

The answer is 7√10−√6

5 0

3 years ago