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3 years ago

The fence along the length of a ranch measures 4,450.5 meters and the fence along the width of a ranch measures 2,890 meters.

1 answer:

4 0

Given:
Length = 4,450.5 meters
Width = 2,890 meters

1 kilometer = 1,000 meters

Length: 4,450.5 meters * 1km/1,000 meters = 4,450.5 km/1,000 = 4.4505km

Width: 2,890 meters * 1km/1,000 meters = 2,890 km/1,000 = 2.89 km

4.4505 km - 2.89 km = 1.5605 km

The length of the fence is 1.5605 km longer than the width of the fence.

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The length of a new rectangular playing field is 3 yards longer than double the width. If the perimeter of the rectangular playi

Romashka-Z-Leto [24]

Answer:

121 x 59 (Dimensions)

Step-by-step explanation:

6x + 6 = 360

Take away 6 from both sides

6x = 354

Divide both sides by 6

x = 59

The length is 2x + 3 because it says 3 more than double the width (lenght is x). You have 2 widths and 2 lengths. The two widths are both x (2x). The two lengths are both 2x + 3 (4x + 6). 2x + 4x + 6 = 6x + 6.

Now input 59 for x

2(59) + 3 = 121

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Check Your Answer

121 is just one length and 59 is just one width, so you have to multiply both sides by 2 because you have 2 widths and 2 lengths.

121 x 2 = 242

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3 years ago

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$\bf \textit{160 at 8:12 ratio}\qquad \qquad \cfrac{8\cdot \frac{160}{8+12}}{12\cdot \frac{160}{8+12}}\implies \cfrac{8(8)}{12(8)}\implies \cfrac{64}{96}$

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Step-by-step explanation:

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A batch consists of 12 defective coils and 88 good ones. Find the probability of getting two good coils when two coils are rando

Alenkinab [10]

Answer:

0.7744 = 77.44% probability of getting two good coils when two coils are randomly selected

Step-by-step explanation:

For each coil, there are only two possible outcomes. Either it is good, or it is not. Since the coil taken is replaced, the probability of choosing a good coil on a trial is independent of any other trial, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

88 out of 100 are good:

This means that $\pi = \frac{88}{100} = 0.88$

Find the probability of getting two good coils when two coils are randomly selected.

This is P(X = 2) when n = 2. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,2}.(0.88)^{2}.(0.12)^{0} = 0.7744$

0.7744 = 77.44% probability of getting two good coils when two coils are randomly selected

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3 years ago