Gabe likes to draw monsters. Each monster he draws had 8 eyes he draws 4 monsters. How many eyes do the monsters have in all ?

Mathematics

1 answer:

Eduardwww [97]4 years ago

5 0

Answer:

The answer is 32

Step-by-step explanation:

8x4=32

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At 6,000 calories a day, a person with genotype X will weigh 285 lbs with much energy and high cholesterol levels. At 3,500 calo

kkurt [141]

Answer: norm of reaction

Step-by-step explanation:

norm of a reaction which is the capacity of a genotype to produce different phenotypes in response to the environment. The group of phenotypic expressions under different environmental conditions. A organism can own all the genetic complements necessary to fully development. However, if there is no suitable nutricion, the development does not occur. The environment only changes the phenotypic characteristics within certain limits ruled by the genotype (norm of reaction). The genotype do not determine a phenotypes, but a amount of possible phenotypes, a norm of reaction. A certain phenotype will materialize itself depending on the interaction of a certain genotype with the environment.

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3 years ago

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$