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2 years ago

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Joshua uses his thermometer and finds the boiling point of ethyl alcohol to be 75o C. He looks in a reference book and finds tha

t the actual boiling point of ethyl alcohol is 80oC. What is his percent error?

2 answers:

Crazy boy [7]2 years ago

5 0

Answer:

the percent error is 6.25%

Step-by-step explanation:

The computation of the percent error is shown below:

Percent error is

= (Experimental value - accepted value) ÷ (experimental value) × 100

= (80 degrees - 75 degrees) ÷ (80 degrees) × 100

= (5 degrees) ÷ (80 degrees) × 100

= 0.0625 × 100

= 6.25%

hence, the percent error is 6.25%

We simply applied the above formula so that the correct value could come

And, the same is to be considered

Mekhanik [1.2K]2 years ago

5 0

Answer:

6.25

Step-by-step explanation:

I just did this

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If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)

Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

$\log_dabc=\log_da+\log_db+\log_dc$

Then using the change of base formula, we can derive the relationship

$\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}$

This immediately tells us that

$\log_dc=\dfrac1{\log_cd}=\dfrac12$

Notice that none of $a,b,c,d$ can be equal to 1. This is because

$\log_1x=y\implies1^{\log_1x}=1^y\implies x=1$

for any choice of $y$ . This means we can safely do the following without worrying about division by 0.

$\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}$

so that

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

Similarly,

$\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}$

so that

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

Another way to do this:

$\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}$

$\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}$

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

Then

$abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}$

So we have

$\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}$

4 0

2 years ago

Determine the equation of the line that passes through the point (7, -23)and is perpendicular to the line y=1/3x-6Enter your ans

marishachu [46]

We have a line y = 1/3x -6

We want a line that is perpendicular to this line

Perpendicular lines have slopes that multiply to -1

1/3 * m = -1

3 * 1/3 *m = -1 * 3

m = -3

The slope of the perpendicular line is -3

y = mx+b where m is the slope and b is the y intercept

y = -3x+b

We have a point on the line ( 7 ,-23)

Substitute this point into the equation

-23 = -3(7)+b

-23 = -21+b

Add 21 to each side

-23+21 = -21+21+b

-2 = b

y = -3x-2

In slope intercept form, the line perpendicular passing through (7,-23) is

y = -3x-2

3 0

10 months ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Rus_ich [418]

Answer:

Diverges

Step-by-step explanation:

We can solve this by using integral by parts:

Let

$u=lnx$ $dv=(1/x)dx$

$du=(1/x)dx$ $v=lnx$

$\int\limits {lnx/x} \, dx = (lnx)^2-\int\limits {lnx/x} \, dx$

We can add

$\int\limits {lnx/x} \, dx$ to both sides

$\int\limits{lnx/x} \, dx = (1/2)\cdot{(lnx)^2}$

We can evaluate the limit between 2 and infinity.

If x tends to infinity the limit will be infinity and therefore the integral diverges to ∞

6 0

2 years ago

A population of mice is increasing exponentially. On Monday there were 120 most. One month later there were 132 most. Ready func

djyliett [7]

Answer:

$y=120 * 1.1^x$ --- function

The monthly rate is 10%

Step-by-step explanation:

Given

Let

$x \to months$

$y \to mice$

So, we have:

$(x_1,y_1) = (0,120)$ --- Monday

$(x_2,y_2) = (1,132)$ --- One month later

Required

The function

The function is represented as:

$y=ab^x$

In $(x_1,y_1) = (0,120)$ , we have:

$120 = a * b^0$

$120 = a * 1$

$120 = a$

$a=120$

In $(x_2,y_2) = (1,132)$ , we have:

$132 = a * b^1$

$132 = a * b$

Substitute: $a=120$

$132 = 120 * b$

Solve for b

$b = \frac{132}{120}$

$b = 1.1$

So, the function is:

$y=ab^x$

$y=120 * 1.1^x$

To calculate the monthly rate (r), we have:

$y =a(1 + r)^x$

Compare to: $y =ab^x$

$1 + r = b$

Make r the subject

$r = b-1$

Substitute $b = 1.1$

$r = 1.1-1$

$r = 0.1$

Express as percentage

$r = 0.1*100\%$

$r = 10\%$

5 0

2 years ago

HERE IS THE ONE WITH THE QUESTION LOL PLZ ANSWER

ZanzabumX [31]

Answer:

The graph goes with the third chart

Step-by-step explanation:

Y starts at 3 and then goes up to 4,5, and then 6 while x starts at 0 and goes up to 1,2, and 3

8 0

2 years ago