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Leviafan [203]
2 years ago
7

Here are the ingredients for making fish pie for 6 people.

Mathematics
1 answer:
yarga [219]2 years ago
6 0
Clare would need 80 of fish, (1/3 of 240 is 80.)
Ian would need 200g of butter (If you find the amount of each ingredient needed for a one person-pie, and multiply them by the amount of people you need to feed, you will get your answer.)
We know that 80g of butter is used for 6 people, so by dividing 80 by 6, we know that for a one person-pie, you would need 13.33g of butter. Multiply 13.33 by 15 to find out how many grams of butter he would need for a pie for 15 people.
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Traveling carnivals move from town to town, staying for a limited number of days before moving to the next stop. The management
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Three times the sum of half Carlita’s age and 3 is at least 12.What values represent Carlita’s possible age? graph the solution
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<em><u>3 x 12 </u></em>

<em><u>I think because you find out how to graph in the corresponding area on the graph </u></em>

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The base of pyramid A is a rectangle with a length of 10 meters and a width of 20 meters. The base of pyramid B is a square with
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Given:
Pyramid A: Base is rectangle with length of 10 meters and width of 20 meters.
Pyramid B: Base is square with 10 meter sides.
Heights are the same.

Volume of rectangular pyramid = (L * W * H) / 3
Volume of square pyramid = a² * h/3

Let us assume that the height is 10 meters.
V of rectangular pyramid = (10m * 20m * 10m)/3 = 2000/3 = 666.67 m³
V of square pyramid = (10m)² * 10/3 = 100m² * 3.33 = 333.33 m³

The volume of pyramid A is TWICE the volume of pyramid B.

If the height of pyramid B increases to twice the of pyramid A, (from 10m to 20m),  

V of square pyramid = (10m)² * (10*2)/3 = 100m² * 20m/3 = 100m² * 6.67m = 666.67 m³

The new volume of pyramid B is EQUAL to the volume of pyramid A.
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An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results an
kotegsom [21]

Answer:

The mean number of successful surgeries is 1.57.

The variance of the number of successful surgeries is 0.3111.

Step-by-step explanation:

STEP 1

If the tear on the left knee has a rim width of less than 3mm, the probability that the surgery on the left knee will be successful (ls) is 0.90.

That isP(Is)=0.90

.

The probability that the surgery on the left knee will fail (lf) is 0.10. That isP(V)=0.10

.

If the tear on the right knee has a rim width of 3-6 mm, the probability that the surgery on the right knee will be successful (rs) is 0.67. That isP(rs)=0.67

.

The probability that the surgery on the right knee will fail (rf) is 0.33.

That isP(vf)= 0.33

.

Let the random variable X denote the number of successful surgeries. The range of X is{0,1,2)

.

Now find the probabilities associated with the possible values of X. The number of successful surgeries is equal to 0 if the surgeries on both knees fail. Since the surgeries are independent, we have:

P(X = 0)=P(lf and rf)

= P()P(rf)

=(0.10)(0.33)

= 0.033

The number of successful surgeries is equal to 1 if the surgery on one knee is successful and the surgery on the other knee fails, that is

P(X =1)= P((ls and rf) or (if and rs))

= P(Is)P(rf)+P(V)P(rs)

= (0.90)(0.33)+(0.10)(0.67)

= 0.364

The number of successful surgeries is equal to 2 if the surgeries on both knees are successful, that is

P(X = 2)=P(ls and rs)

= P(Is) P(rs)

=(0.90)(0.67)

= 0.603

So, the probability mass function of X is the following

X        0              1               2

f(x)     0.033      0.364      0.603

The mean number of successful surgeries is,

E(X)=∑xP(x)

=(0×0.033) +(1×0.364)+(2×0.603)

=1.57

the mean number of successful surgeries is 1.57

The expected value is obtained by taking the summation of the product of each possible value of the random variable X with its corresponding probabilities. Thus, the mean number of successful surgeries is 1.57.

STEP 2

The variance of the number of successful surgeries is,

Var(X)=∑x²P(x)-(E(x²)

=(0×0.033) + (1 × 0.364) +(2 × 0.603) - (1.57)²

= 2.776 -(1.57)²

=0.3111

The variance of the number of successful surgeries is 0.3111.

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