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 $\bar X$ = the average length of rods in a randomly selected bundle of steel rods

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)