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3 years ago

Find the distance between each pair of points. (-1, -2), (1,5)

Mathematics

1 answer:

fgiga [73]3 years ago

3 0

Answer:

$\sqrt{53}$

Step-by-step explanation:

To find the distance between two points, we can use the distance formula. The distance formula is:

$\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

By plugging in (-1, -2) and (1, 5) into the distance formula, we get:

$\sqrt{(1-(-1))^{2}+(5-(-2))^{2}}=\sqrt{(1+1)^{2}+(5+2)^{2}}=\sqrt{(2)^{2}+(7)^{2}}=\sqrt{4+49}=\sqrt{53}$

So the distance between the two points is $\sqrt{53}$ .

I hope you find my answer and explanation to be helpful. Happy studying.

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I NEED HELP (I’m sooo dumb lol)

statuscvo [17]

To make a box and whisker plot, first you write down all of the numbers from least to greatest.

0, 1, 3, 4, 7, 8, 10

The median is 4, so that’s the middle line of the plot.

So now we have:
0, 1, 3, [4,] 7, 8, 10

So next we have to find the 1st and 3rd interquartiles..
0, [1,] 3, [4,] 7, [8,] 10

Those are the next 2 points you put on the plot.

Lastly, the upper and lower extremes. These are the highest and lowest numbers in the data.
[0,] 1, 3, 4, 7, 8, [10]

These are the final points on the plot.

To make the box of a box-and-whisker plot, you plot the 3 Medians of the data: 1, 4, and 8, and connect those to make a box that has a line in the middle at 4.

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7 0

3 years ago

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

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3 years ago

The following polynomial can be factored using the techniques of this lesson. x2 + 16

Nat2105 [25]

Answer:

The answer here would be false

Step-by-step explanation:

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mezya [45]

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3 years ago

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MArishka [77]

Answer:

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3 years ago

Read 2 more answers

Find the distance between each pair of points. (-1, -2), (1,5)​

Find the distance between each pair of points. (-1, -2), (1,5)