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2 years ago

Please help me for solve x

Mathematics

1 answer:

dlinn [17]2 years ago

3 0

Answer:

X =6

Step-by-step explanation:

Angles on a straight line add up to 180°

180-110=70

The 70° is the same on the angles above

70+8+17x=180

180-70=110

110-8=102

102/17=6

X=6

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Step-by-step explanation:

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2 years ago

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3 years ago

Read 2 more answers

The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged

Temka [501]

Answer:

a) 0.283 or 28.3%

b) 0.130 or 13%

c) 0.4 or 40%

d) 30.6 mm

Step-by-step explanation:

z-score of a single left atrial diameter value of healthy children can be calculated as:

z= $\frac{X-M}{s}$ where

X is the left atrial diameter value we are looking for its z-score

M is the mean left atrial diameter of healthy children (26.7 mm)

s is the standard deviation (4.7 mm)

Then

a) proportion of healthy children who have left atrial diameters less than 24 mm

=P(z<z*) where z* is the z-score of 24 mm

z*= $\frac{24-26.7}{4.7}$ ≈ −0.574

And P(z<−0.574)=0.283

b) proportion of healthy children who have left atrial diameters greater than 32 mm

= P(z>z*) = 1-P(z<z*) where z* is the z-score of 32 mm

z*= $\frac{32-26.7}{4.7}$ ≈ 1.128

1-P(z<1.128)=0.8703=0.130

c) proportion of healthy children have left atrial diameters between 25 and 30 mm

=P(z(25)<z<z(30)) where z(25), z(30) are the z-scores of 25 and 30 mm

z(30)= $\frac{30-26.7}{4.7}$ ≈ 0.702

z(25)= $\frac{25-26.7}{4.7}$ ≈ −0.362

P(z<0.702)=0.7587

P(z<−0.362)=0.3587

Then P(z(25)<z<z(30)) =0.7587 - 0.3587 =0.4

d) to find the value for which only about 20% have a larger left atrial diameter, we assume

P(z>z*)=0.2 or 20% where z* is the z-score of the value we are looking for.

Then P(z<z*)=0.8 and z*=0.84. That is

0.84= $\frac{X-26.7}{4.7}$ solving this equation for X we get X=30.648

5 0

3 years ago

use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 ), (15 comma

lyudmila [28]

Answer:

volume V of the solid

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

Step-by-step explanation:

The situation is depicted in the picture attached

(see picture)

First, we divide the segment [0, 5] on the X-axis into n equal parts of length 5/n each

[0, 5/n], [5/n, 2(5/n)], [2(5/n), 3(5/n)],..., [(n-1)(5/n), 5]

Now, we slice our solid into n slices.

Each slice is a quarter of cylinder 5/n thick and has a radius of

-k(5/n) + 5 for each k = 1,2,..., n (see picture)

So the volume of each slice is

$\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}$

for k=1,2,..., n

We then add up the volumes of all these slices

$\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}$

Notice that the last term of the sum vanishes. After making up the expression a little, we get

$\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2$

But

$\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)$

we also know that

$\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}$

and

$\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}$

so we have, after replacing and simplifying, the sum of the slices equals

$\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)=\\\\=\displaystyle\frac{5\pi}{4n}\left(\displaystyle\frac{25}{n^2}.\displaystyle\frac{n(n-1)(2n-1)}{6}-\displaystyle\frac{50}{n}.\displaystyle\frac{n(n-1)}{2}+25(n-1)\right)=\\\\=\displaystyle\frac{125\pi}{24}.\displaystyle\frac{n(n-1)(2n-1)}{n^3}$

Now we take the limit when n tends to infinite (the slices get thinner and thinner)

$\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}$

and the volume V of our solid is

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

3 0

3 years ago