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3 years ago

Two hundred ten million,sixty four thousand , fifty in standard form

Mathematics

2 answers:

Vlad1618 [11]3 years ago

6 0

210,064,050 is the correct answer

astra-53 [7]3 years ago

6 0

210,064,050 is the number.

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If 1/3 of a set of numbers is 8 how many numbers are in the total set ?

Eduardwww [97]

Answer:

Step-by-step explanation:

1/3 = 8

3/3 = 8*3

7 0

3 years ago

How many quarts of pure antifreeze must be added to 5 quarts of a 40% antifreeze solution to obtain a 50% antifreeze solution

GarryVolchara [31]

Answer:

2 quarts of 100% antifreeze.

Step-by-step explanation:

5 quarts/ .40 * x/ 1

.40x= 5

x=2

6 0

3 years ago

PLEASE HELP ASAP

Volgvan

The length of each side of the polygon is the number of units on each side of the polygon, and the perimeter of the factory is 1000 feet

<h3>The length of each side of the polygon?</h3>

To do this, we simply calculate the number of units on each side of the polygon.

So, we have:

S1 = 5 units

S2 = 8 units

N1 = 13 units

E1 = 7 units

W1 = 3 units

W2 = 4 units

<h3>The perimeter of the polygon</h3>

This is the sum of the side lengths.

So, we have:

Perimeter = 5 + 8 + 13 + 7 + 3 + 4

Perimeter = 40

Each unit equals 25 feet.

So, we have:

Perimeter = 40 * 25 feet

Evaluate

Perimeter = 1000 feet

Hence, the perimeter of the factor is 1000 feet

Read more about perimeter at:

brainly.com/question/24571594

#SPJ2

8 0

2 years ago

Read 2 more answers

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

b) $P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452$

c) $P(X \geq 8) = 1-P(X$

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

$P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465$

$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

$P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298$

$P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452$

c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago

A contractor is calculating the number of decorative stones for a new rectangular patio. The bids always include a count for the

babunello [35]

Answer:

The number of edge stones needed has a constant rate of change, but the number of stones for the center does not.

Step-by-step explanation:

The complete question is shown in the picture attached below.

Two functions E(x) and C(x) are given in the table along with some of the function values. We have to identify if any of these functions show constant rate of change or not.

By constant rate of change we mean that the slope is constant or in other words, a linear relationship is shown by the function. For a Linear function, the first differences of the function values are same.

By first differences we mean the difference between two consecutive output values. We can see that difference between consecutive input values is constant i.e. 1. If the difference between consecutive output values of any of the functions is same, then that function will be a Linear Function and, therefore, the rate of change for that function will be constant.

Lets analyze E(x) first. The output values are:

34, 52, 70 and 88

The difference between consecutive output values is:

18, 18 and 18

Since, this difference is constant, we can conclude that E(x) is a Linear function with a constant rate of change.

Now lets analyze C(x). The output values are:

62, 133, 260 and 435

The difference between consecutive output values is:

71, 127 and 175

Since, the differences are not constant, C(x) is not a Linear Function and , therefore, the rate of change of C(x) is not constant.

Conclusion:

The number of edge stones which is represented by E(x) has a constant rate of change, but the number of stones for center which is represented by C(x) does not have constant rate of change. This makes the first option our correct answer.

3 0

3 years ago