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4 years ago

5

Find the distance between the given points.

1 answer:

shutvik [7]4 years ago

7 0

Answer:

The correct answer is last option

10

Step-by-step explanation:

Points to remember

<u>Distance formula</u>

Length of a line segment with end points (x1, y1) and (x2, y2) is given by,

Distance = √[(x2 - x1)² + (y2 - y1)²]

<u>To find the distance between given points</u>

Here (x1, y1) = R(0, 0) and (x2, y2) = S(6, 8)

Distance, RS = √[(x2 - x1)² + (y2 - y1)²]

= √[(6 - 0)² + (8 - 0)²]

= √[6² + 8²]

= √[36 + 64]

= √100

= 10

Therefore the correct answer is last option 10

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Find the volume (in terms of pi) of a sphere of its surface area is 400 pi ft2

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3 years ago

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Find x when y = 32, given that x varies inversely as y, and x = 132 when y = 4.

gizmo_the_mogwai [7]

Answer:

When y=32, x=16.5

Step-by-step explanation:

Find x when y = 32, given that x varies inversely as y, and x = 132 when y = 4.

We are given:

x varies inversely with y

We can write it as: $x\:\alpha \:\frac{1}{y}$

$x=\frac{k}{y}$

We have x = 132, when y=4

We can find value of k by using these values

$x=\frac{k}{y}\\132=\frac{k}{4}\\k=132*4\\k=528$

We need to find x when y=32

$x=\frac{k}{y}\\x=\frac{528}{32}\\x=16.5$

So, when y=32, x=16.5

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3 years ago

Principal $400 Rate 5% time 1 year

zheka24 [161]

Answer:400 multiplied by 5 percent then later multiplied by one, Answer: 20$

Step-by-step explanation:

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3 years ago

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The standard deviation of test scores on a certain achievement test is 11.3. If a randomsample of 81 students had a sample mean

Svetradugi [14.3K]

Answer: (72.535, 76.665)

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

, where $\sigma$ = population standard deviation.

n= Sample size

$\overline{x}$ = sample mean.

$z^*$ = Critical z-value (two-tailed)

Given : $\sigma=11.3$

n= 81

$\overline{x}=74.6$

The critical values for 90% confidence interval : $z^*=\pm1.645$

Now, the confidence interval for population mean would be :

$74.6\pm (1.645)\dfrac{11.3}{\sqrt{81}}\\\\=74.6\pm(1.645)(\dfrac{11.3}{9})\approx74.6\pm2.065\\\\=(74.6-2.065,\ 74.6+2.065)\\\\=(72.535,\ 76.665)$

Hence, a 90 percent confidence interval estimatefor the average score of all students. : (72.535, 76.665)

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3 years ago