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Alona [7]
3 years ago
14

Mrs.Cranston bought five bottles of water for $0.90 each and 8 pounds of meat. She paid a total of $26.90 for these items, not i

ncluding tax. What was the price per pound of the meat?
Mathematics
1 answer:
Morgarella [4.7K]3 years ago
6 0
Water = 5 bottles * $0.90 = $4.50
meat = 8 pounds * x amount.

Total = water + meat prices = $26.90
26.90 = $4.50 + 8x
22.40 = 8x
2.8 = x

The price per pound of meat is $2.80.
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ser-zykov [4K]
<span>480/1 hope this help</span>
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4 years ago
Suppose you wish to test if a number cube (die) is loaded or not. If the die is not loaded, the theoretical probabilities for ea
Sonja [21]

Answer:

Goodness of fit

Step-by-step explanation:

Given

The theoretical probabilities

<em>See comment for complete question</em>

<em></em>

Required

The type of test to be use

From the question, we understand that you are to test if the die is loaded or not using the given theoretical probabilities.

This test can be carried out using goodness of fit test because the goodness of fit is basically used to check the possibility of getting the outcome variable from a distribution. In this case, the outcome of the variables are the given theoretical probabilities.

In a nutshell, the goodness fit of test determines if the given data (in this case, the theoretical probabilities) is a reflection of what to expect in the original population.

3 0
3 years ago
Consider in the figure below.
yawa3891 [41]

Here BD is perpendicular bisector

So

  • UB=BV=74

\\ \tt\hookrightarrow UV=74+74=148

Apply Pythagorean theorem

\\ \tt\hookrightarrow BD^2=UD^2-UB^2=UD^2-VD^2

  • UD=VD=78=TD

\\ \tt\hookrightarrow TC^2=TD^2-CD^2

\\ \tt\hookrightarrow TC^2=78^2-30^2

\\ \tt\hookrightarrow TC^2=6084-900=5184

\\ \tt\hookrightarrow TC=72

6 0
3 years ago
Read 2 more answers
Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2
Verdich [7]

Answer:

(a) f'(x)=-\frac{2}{x^3}

(b) y=-0.25x+0.75

Step-by-step explanation:

The given function is

f(x)=\frac{1}{x^2}                  .... (1)

According to the first principle of the derivative,

f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}

Cancel out common factors.

f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}

By applying limit, we get

f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}

f'(x)=\frac{-2x)}{x^4}

f'(x)=\frac{-2)}{x^3}                         .... (2)

Therefore f'(x)=-\frac{2}{x^3}.

(b)

Put x=2, to find the y-coordinate of point of tangency.

f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}

Substitute x=2 in equation 2.

f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

y-y_1=m(x-x_1)

y-0.25=-0.25(x-2)

y-0.25=-0.25x+0.5

y=-0.25x+0.5+0.25

y=-0.25x+0.75

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0
3 years ago
5 is what percent of 23
Eddi Din [679]
5 = what percent of 23

5 = x% of 23

5 = (x/100)*23

5 = 23x/100

5*100 = 23x

23x = 5*100

x = 5*100/23

x = 21.739

5 is ≈ 21.739% of 23 
8 0
3 years ago
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