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3 years ago

What is 1/9 - 2 4/9 - 5/9

Mathematics

2 answers:

ANEK [815]3 years ago

4 0

Answer:

Step-by-step explanation:

Ymorist [56]3 years ago

3 0

1/9-22/9-5/9
-21/9-5/9
-26/9

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In a study relating the degree of warping, in mm, of a copper plate (y) to temperature in °C (x), the following summary statisti

elena-14-01-66 [18.8K]

Answer:

y = 0.3035 + 0.0082x

0.6315 mm

x = 23.9634 °C

Step-by-step explanation:

a. Compute the least-squares line for predicting warping from temperature. Round the answers to four decimal places.

We need to find an equation of the form

y = b + mx

where m is the slope and b the Y-intercept.

The slope m can be computed with the formula

$\bf m=\displaystyle\frac{\displaystyle\sum_{i=1}^n(x_i-\bar x)(y_i-\bar y)}{\displaystyle\sum_{i=1}^n(x_i-\bar x)^2}$

Replacing the values in our formula (we will round at the end of the calculations)

$\bf m=\displaystyle\frac{806.94}{98,775}=0.008169476$

the Y-intercept b is computed with the formula

$\bf b=\bar y-m\bar x$

therefore we have

$\bf b=0.5188-0.008169476*26.36=0.303452613$

and the least-squares line rounded to 4 decimals would be

y = 0.3035 + 0.0082x

b. Predict the warping at a temperature of 40°C. Round the answer to three decimal places.

We simply replace x with 40 to get

y = 0.3035 + 0.0082*40 = 0.6315 mm

c. At what temperature will we predict the warping to be 0.5 mm? Round the answer to two decimal places

Here, we replace y with 0.5 and solve for x

0.5 = 0.3035 + 0.0082x ===> x = (0.5-0.3035)/0.0082 ===>

x = 23.9634 °C

6 0

3 years ago

What is 8/10 simplified by

Oksanka [162]

Answer:

4/5

Step-by-step explanation:

that is it

6 0

2 years ago

Read 2 more answers

Both the football and volleyball teams have games today. The football team plays every 7 days. The volleyball team plays every 3

Allisa [31]

They will have games on the same day in 21 days

6 0

3 years ago

If Besty has 14 t-shirts and 8 pairs of pants, the ratio of pants to t-shirts is?

vlabodo [156]

Answer:

4:7

Step-by-step explanation:

8 : 14

8/2 : 14/2

= 4 : 7

6 0

3 years ago

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

3 years ago