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3 years ago

Please help me understand this problem! I don’t understand the methods to understand how to solve it

Mathematics

1 answer:

Doss [256]3 years ago

7 0

Answer:

Perimeter at the big rectangle is 156 cm.

Step-by-step explanation:

Let's see how to calculate it.

1. First of all you know that perimeter in the blue one is 20cm, so imagine this:

L (long side); S (short side)

2L + 2S =20

and we consider that L = 4S

So, solving the equation:

2.4S + 2S =20

10S=20

S=20/10

S=2

L=8

2. Side at the gold square is 8, the same at the long side in the blue rectangle. So, if you see on the right side in the big one, we got 2 + 8 + 8 + (?). Take a look to the green. Green square is the gold + a short piece and you can understand the short piece as 2 short sides from the blue. If we give numbers we have 8 + 2 + 2, 12.

Now, 2+8+8+12 = 30cm

3. Let's go to the long side in the big one.

We have long side from blue (8) and as you see, side at the orange square must be side at the yellow + short at the blue, so 8+2 =10. We have four oranges square so 10+10+10+10=40, and +8 =48

4. Now that we have the two sides in the big one, let's find the perimeter with the rectangle formula:

2L + 2S =P

2.48 + 2.30 = 156 cm.

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Answer:

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Step-by-step explanation:

The model is a right triangle with ladder being the hypotenuse and the angle between the ground and the ladder is 70°

Using the cosine ratio, with l being the length of the ladder.

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Step-by-step explanation:

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Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65

erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

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

where, $\mu$ = population mean = 2.65

$\sigma$ = standard deviation = 0.85

n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

P(X bar <= 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z < 2.06) = 0.98030

P(X bar <= 2.65) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .

8 0

3 years ago