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3 years ago

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Enrique is measuring the mass of a flower petal. Which unit of measurement should he use to eliminate the need to write the valu

e in scientific notation? grams. centimeters. liters. kilograms.

2 answers:

GuDViN [60]3 years ago

6 0

Answer:

do you have a pic of it.

Step-by-step explanation:

alukav5142 [94]3 years ago

5 0

Answer:

Grams

Step-by-step explanation:

Edg. 2020 :D

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PLEASE PLEASE OLEASE HELP ME ASAP I WILL LOVE U FOREVER

vodomira [7]

Answer:

m=3/2

look how many units go up and to the side between 2 points

As seen here, we go up 3 units and over 2 units from (0, -3) to (2, 0.) Therefore, slope is 3/2

Step-by-step explanation:

8 0

3 years ago

Janelle draws line segments AB and BC in the same plane such that AB = 8 cm and BC = 6 cm. Then Janelle draws

Dovator [93]

Answer:

Step-by-step explanation:

• AB, BC, and AC form a triangle. Enter a possible value of AC....

So it asks for only a possible value of AC as there are many possible values.

Given AB = 8 cm and BC = 6 cm, they are in the ratio of 3:4.

Line segments of 3, 4 and 5 length will form a right-angled triange.

A possible value of AC = 5*2 = 10cm

• Points A, B, and C lie on the same line, and C lies between A and B.

So AC+CB = AB

AC+6 = 8

AC = 2cm

Enter this value of AC in the second

response box.

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3 years ago

Read 2 more answers

A total of 26 bills are in a cash box. Some of the bills are one-dollar bills, and the rest are five-dollar bills. The total amo

kobusy [5.1K]

Answer:

Number of 1 $ bill = 20

Number of 5 $ bill = 6

Explanation:

Let p represent number of one dollar bolls and q represent number of five bills.

A total of 26 bills are in a cash box.

p + q = 26 ------------------------- Equation 1

The total amount of cash in the box is $50

p + 5q = 50 ------------------------- Equation 2

Equation 2 - Equation 1,

p + 5q - p -q = 50 - 26

4q = 24

q = 6

Substituting in equation 1.

p + 6 = 26

p = 20

Number of 1 $ bill = 20

Number of 5 $ bill = 6

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3 years ago

Use the expression in the accompanying discussion of sample size to find the size of each sample if you want to estimate the dif

rodikova [14]

Answer:

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3 years ago

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).

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

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

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