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2 years ago

S(-4,-6) and T (-7,-3); Find R

Mathematics

1 answer:

Illusion [34]2 years ago

4 0

Answer:

what formula are you using? and do you have a picture of it?

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The difference between two positive integers is 3. If the smaller is added to the square of the larger, the sum is 87.

Komok [63]

Let x be larger integer so x-3 is the smaller
so x^2+ x-3 = 87
x^2 +x -90=0 and factor
(x-9)(x+10)=0
x=9 or x=-10 but x must be positive
so x=9 and smaller number is 6

6 0

3 years ago

I need the answers fast !!! this test is due today at 11:59

zhenek [66]

Answer:

Options 2 and 6

Step-by-step explanation:

Those are the locations of the points on the graph. If you look closely, you will see that it counts by twos.

5 0

2 years ago

What number is equivalent to 3/4 b-2/3 b =-9

Shtirlitz [24]

Hi,

$b = - 9 \\ \frac{3}{4} \times b - \frac{2}{3} \\ 0.75 \times - 9 - 0.67 \\ - 6.75 - 0.67 = - 7.42$

Answer: -7.42

Hope this helps.
r3t40

5 0

3 years ago

For a one-tailed test with a = .05 and a sample of n = 9, the critical value for the t statistic is t = 1.860.

ahrayia [7]

I think its B. False if im wrong sorry hope this helped tho :p

6 0

3 years ago

Listed below are speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, California (based on data from SigAlert

adell [148]

Answer:

Option (B) is the correct answer to the following question.

Step-by-step explanation:

Step-1: We have to find the Mean of the series.

The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.

$Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}$ $= 60.67$

Step-2: We have to find the Standard Deviation.

Let Standard Deviation be x.

Formula of Standard Deviation is: $s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}$

Put value in formula of Standard Deviation,

$s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}$ = 40.75

Step-3: Then, we have to find the critical value by chi-square.

$X_{1-\alpha/2}^{2}=3.82$

$X_{1-\alpha/2}^{2}=21.92$

Then, find the confidence interval which is 95%.

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9$

$\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9$

5 0

3 years ago