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weqwewe [10]
3 years ago
14

Need Help With This One Please!!!!

Mathematics
2 answers:
Flauer [41]3 years ago
6 0

For this case we have that by definition, the area of a triangle is given by:

A = \frac {1} {2} b * h

Where:

b: It's the base

h: It's the height

Substituting the values, according to the given data:

225 = \frac {1} {2} (15) * h\\225 = 7.5h

Dividing between 7.5 on both sides of the equation:

h = \frac {225} {7.5}\\h = 30

Thus, the height of the triangle is 30ft

Answer:

30ft

horrorfan [7]3 years ago
6 0

Answer: 30 feet.

Step-by-step explanation:

The formula used to calculate the area of a triangle is:

A=\frac{bh}{2}

Where "b" is the base and "h" is the height.

In this case you know that the area of this triangle is 225 ft² and its base is 15 ft. Then, you can substitute these values into the formula A=\frac{bh}{2} and solve for the height "h":

225ft^2=\frac{(15ft)h}{2}

(2)(225ft^2)=(15ft)h

h=\frac{(2)(225ft^2)}{15ft}

h=30ft

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Step-by-step explanation:

Step one:

given data

We are told that the unit cost per pound is $5.89

Required

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5 0
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A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints
Sliva [168]

Answer:

a)0.099834

b) 0

Step-by-step explanation:

To solve for this question we would be using , z.score formula.

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.

a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.

Standard Deviation = √variance

= √0.16 = 0.4

Standard deviation = 0.4

Mean = 21.37

x = 20.857

z = (x-μ)/σ

z = 20.857 - 21.37/0.4

z = -1.2825

P-value from Z-Table:

P(x<20.857) = 0.099834

b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.

z score formula used = (x-μ)/σ/√n

x = 20.857

Standard deviation = 0.4

Mean = 21.37

n = 100

z = 20.857 - 21.37/0.4/√100

= 20.857 - 21.37/ 0.4/10

= 20.857 - 21.37/ 0.04

= -12.825

P-value from Z-Table:

P(x<20.857) = 0

c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.

5 0
3 years ago
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