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1 year ago

Find 100% of the number 41 is 0.30% of ____ hurry please

Mathematics

2 answers:

Marina CMI [18]1 year ago

5 0

Answer:

if 41 is 100% of 41, then divide 41 by 0.30 to get 136.7

Step-by-step explanation:

lakkis [162]1 year ago

4 0

Answer:

Step-by-step explanation:

Let the number be 'x'

0.30% of x = 41

$\sf \dfrac{0.30}{100}*x = 41\\\\ x = 41*\dfrac{100}{0.3}$

x = 13666

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1/3 of the fish in the tank are snappers. 2/9 of the fish are catfish and the rest are goldfish. There are 40 goldfish in the ta

Paraphin [41]

<u>Find fraction of snappers and catfish:</u>

1/3 + 2/9 = 3/9 + 2/9 = 5/9

<u>Find fraction of goldfish:</u>

1 - 5/9 = 4/9

4/9 = 40 goldfish

1/9 = 40 ÷ 4 = 10

9/9 = 10 x 9 = 90

Answer: 90 fish in the tank

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3 years ago

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A clinical test on humans of a new drug is normally done in three phases. Phase I is conducted with a relatively small number of

navik [9.2K]

Answer:

A clinical test on humans of a new drug is normally done in three phases. Phase I is conducted with a relatively small number of healthy volunteers. For example, a phase I test of a specific drug involved only 7 subjects.

Step-by-step explanation:

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2 years ago

A hardware store owner is replenishing his stock of base paint for an upcoming home improvement sale. He conducted a survey to d

antiseptic1488 [7]

Solution:

Matte Satin Glossy Total

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Answer: Percentage of contractors who prefer the glossy finish is:

$\frac{0.18}{0.48}=0.375
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Therefore, the option D. 37.5% is correct

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2 years ago

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Pls help me calculate the total area of compound shapes?!?!?

Elza [17]

Answer:

Left: 42ft^2

Right: 21ft^2

Step-by-step explanation:

Its the same question as last time.

Left: (3*10) + (2*6) = 42

Right: (3*5) + (2*3) = 21

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1 year ago

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

7 0

2 years ago