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2 years ago

What is the answer for this?

Mathematics

1 answer:

andreev551 [17]2 years ago

4 0

Whats the equation i dont think i can solve with out an equation

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Jessica’s friend lent her $5. Later that day Jessica gave her friend back $1.75 Which rational number represents the overall cha

Vesna [10]

Answer:

-$3.25

Step-by-step explanation:

She had $5 to begin with and ended with $1.75 the change is - because she has less 5-1.75=3.25

3 0

2 years ago

Read 2 more answers

A right circular cylinder has a volume of 1 cubic meter and a height of 0.6 meters what is the radius of the cylinder to the nea

lapo4ka [179]

The volume formula of a cylinder is :

$V=\pi r^2h$

From the problem, the volume is 1 m^3 and the height is 0.6 m.

Substitute the given to the formula :

$\begin{gathered} 1=\pi r^2(0.6) \\ 1=3.14(0.6)r^2 \\ r^2=0.53 \end{gathered}$

Take the square root of both sides of the equation :

$\begin{gathered} r^2=0.53 \\ r=\sqrt[]{0.53} \\ r=0.728 \end{gathered}$

The answer is r = 0.728 m

8 0

9 months ago

5 friends split 1/2 of a chocolate bar. How much does each friend eat? Draw a model.

Vinvika [58]

Answer:

1/10

Step-by-step explanation:

the friends eat 1/10 of the whole bar or 1/5 of half a bar

3 0

2 years ago

Perform the indicated operation. 5 2/3 ÷ 2 1/8

bezimeni [28]

5 2/3 ÷ 2 1/8 = 17/3 ÷ 17/8 = 17/3 x 8/17 = 8/3 = 2 2/3

8 0

2 years ago

The fracture toughness (in ) of a particular steel alloy is known to be normally distributed with a mean of 28.3 and a standard

Effectus [21]

This is a conditional probability question.

The conditional probability is given as follows: The probability of an event A given B is given by

$P(A|B)= \frac{P(A\cap B)}{P(B)}$

Thus, given that the fracture toughness of a particular steel alloy is known to be normally distributed with a mean of 28.3 and a standard deviation of 0.77.

The probability that the fracture toughness will be between 29 and 40.3 given that the fracture toughness is at least 27 is given by

$P(29 \leq X \leq 40.3 \, | \, X \geq 27)= \frac{P(29 \leq X \leq 40.3)\cap P(X \geq 27)}{P(X \geq 27)} \\ \\ =\frac{P(29 \leq X \leq 40.3)}{P(X \geq 27)}$

$P(29 \leq X \leq 40.3)=P\left(z \ \textless \ \frac{40.3-28.3}{0.77}\right)- P\left(z \ \textless \ \frac{29-28.3}{0.77}\right) \\ \\ =P(z\ \textless \ 15.58)-P(z\ \textless \ 0.91)=1-0.81835=0.18165$

$P(X \geq 27)=1-P(x\ \textless \ 27) \\ \\ =1-P\left(z\ \textless \ \frac{27-28.3}{0.77}\right)=1-P(z\ \textless \ -1.688) \\ \\ =1-0.04568=0.95432$

Thus, the probability that the fracture toughness will be between 29 and 40.3 given that the fracture toughness is at least 27 is given by

$\frac{P(29 \leq X \leq 40.3)}{P(X \geq 27)}= \frac{0.18165}{0.95432} =\bold{0.1903}$

6 0

2 years ago