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3 years ago

8

Mr.varela bought a block of fudge that weighed 3/4 of a pound. He cut the fudge into 2 equal pieces. What was the weight of each

piece of fudge?

1 answer:

OleMash [197]3 years ago

6 0

To do this, multiply 3/4 by 1/2 and you end up with 3/8 lb. or 0.375 lb.

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Help with five plz!!!!!

frutty [35]

Answer:

No there cannot be the same number of stickers on each page.

Step-by-step explanation:

If you want to find out how many stickers need to be in every page to be even you would add all the stickers up. 6+6+9+10+11= 42. Take the 42 and divide it by 5 to see how many stickers would go in each page. This will give you 8.4. However since this number is a decimal it can't be split evenly in whole stickers for each page. Meaning that it wouldn't be possible for each page to have a evenly distributed number of stickers per each page.

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3 years ago

Please help me with this question.

Stels [109]

Step-by-step explanation:

Mean of the origibal set = 20

No. of terms = 10

Sum = Mean x No. of terms

= 20 x 10 = 200

The sum of additional numbers to make the new set of observstion is 4 + 8 + 12...... + 40 = 220

New Sum = 200 + 220 = 420

The no of terms stay the same

New Mean = New Sum/ N

= 420/10

= 42

7 0

2 years ago

The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event th

choli [55]

Answer:

Step-by-step explanation:

From the given information:

The uniform distribution can be represented by:

$f_x(x) = \dfrac{1}{1500} ; o \le x \le \ 1500$

The function of the insurance is:

$I(x) = \left \{ {{0, \ \ \ x \le 250} \atop {x -20 , \ \ \ \ \ 250 \le x \le 1500}} \right.$

Hence, the variance of the insurance can also be an account forum.

$Var [I_{(x}) = E [I^2(x)] - [E(I(x)]^2$

here;

$E[I(x)] = \int f_x(x) I (x) \ sx$

$E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx$

$= \dfrac{1}{1500 } \dfrac{(x - 250)^2}{2} \Big |^{1500}_{250}$

$\dfrac{5}{12} \times 1250$

Similarly;

$E[I^2(x)] = \int f_x(x) I^2 (x) \ sx$

$E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250)^2 \ dx$

$= \dfrac{1}{1500 } \dfrac{(x - 250)^3}{3} \Big |^{1500}_{250}$

$\dfrac{5}{18} \times 1250^2$

∴

$Var {I(x)} = 1250^2 \Big [ \dfrac{5}{18} - \dfrac{25}{144}]$

Finally, the standard deviation of the insurance payment is:

$= \sqrt{Var(I(x))}$

$= 1250 \sqrt{\dfrac{5}{48}}$

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2 years ago

What is the following product? Sqr5x^8y^2 sqr10x^3 sqr12y

PSYCHO15rus [73]

Step-by-step explanation:

$\sqrt{5 {x}^{8} {y}^{2} } \times \sqrt{10 {x}^{3} } \times \sqrt{12y} \\ \\ = {x}^{4} y \sqrt{5} \times x \sqrt{10x} \times 2 \sqrt{3y} \\ \\ = 2 {x}^{5} y \sqrt{5 \times 10x \times 3y} \\ \\ = 2 {x}^{5} y \sqrt{5 \times 5 \times 2x \times 3y} \\ \\ = 2 {x}^{5} y \times 5\sqrt{ 2x \times 3y} \\ \\ = 10 {x}^{5} y \sqrt{6xy} \\$

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3 years ago

What function is graphed below?

erik [133]

Because the graph is shifted up three units from the original mother function (f(x) = log x), and therefore, the answer is f(x) = log x + 3.

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3 years ago

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