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AlladinOne [14]
3 years ago
5

How do we solve this problem?

Mathematics
1 answer:
MaRussiya [10]3 years ago
5 0
To solve the problem you must find the area of every square. For F one of the squares has a side length of 3 and because this is a square the other sides are also 3. To find the area we multiply two of the sides so 3 times 3 is 9
You do this with both the other squares
3*3=9
4*4=16
The area of the large square is given as 25
So 9+16=25 which means this is not the correct answer

H.
9*9=81
Given as 144
The area of the large square is 21*21 which is 441
144+81= 225
So because the two small squares DO NOT equal the area of the large square this is the correct answer.
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A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2-cm and a standard dev
Varvara68 [4.7K]

Answer:

The probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is 0.65173.

Step-by-step explanation:

We are given that a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2 cm and a standard deviation of 2.1 cm. For shipment, 17 steel rods are bundled together.

Let \bar X = <u><em>the average length of rods in a randomly selected bundle of steel rods</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean length of rods = 259.2 cm

           \sigma = standard deviaton = 2.1 cm

           n = sample of steel rods = 17

Now, the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is given by = P(\bar X > 259 cm)

 

     P(\bar X > 259 cm) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{259-259.2}{\frac{2.1}{\sqrt{17} } } ) = P(Z > -0.39) = P(Z < 0.39)

                                                                = <u>0.65173</u>

The above probability is calculated by looking at the value of x = 0.39 in the z table which has an area of 0.65173.

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3 years ago
HELPPPP PLZZZZZZZZZZ ITS TIMEDDDDDD Identify the constant and the coefficient in the term below and tell me the difference betwe
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Can you take a picture of the question so I can solve it
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3 years ago
Can someone please<br> Help me?
Mila [183]

9514 1404 393

Answer:

  see below

Step-by-step explanation:

It is easiest to compare the equations when they are written in the same form.

The first set can be written in slope-intercept form.

  y = 2x +7

  y = 2x +7 . . . . add 2x

These equations are <em>identical</em>, so have infinitely many solutions.

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The second set can be written in standard form.

  y +4x = -5

  y +4x = -10

These equations <em>differ only in their constant</em>, so have no solutions.

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The third set is already written in slope-intercept form. The equations have <em>different slopes</em>, so have exactly one solution.

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3 years ago
The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is r
svp [43]

Answer:

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value between c and d is given by:

P(c \leq X \leq d) = \frac{d - c}{b - a}

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min.

This means that a = 50, b = 52

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P(51.5 \leq X \leq 51.7) = \frac{51.7 - 51.5}{52 - 50} = \frac{0.2}{2} = 0.1

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

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Answer:

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Step-by-step explanation:

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