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4 years ago

Can you create a triangle with sides 17ft , 8ft , and 25ft ?

Mathematics

2 answers:

Vlad1618 [11]4 years ago

5 0

Answer:

Step-by-step explanation:

Yes

Igoryamba4 years ago

3 0

Answer:

No.

Step-by-step explanation:

The pythagorean theorem states that a^2 + b^2 = c^2. By the basic structure of a triangle, the hypotenuse is the longest side, c. In this case, 25 ft is the value of c. 17^2 + 8^2 = 353. 25^2 = 625, so this isn't an accurate triangle.

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A machine of a fast food restaurant produces a bag of French fries

Anni [7]

At 08:00 there are 18 bags
At 08:10 there are an extra 10 produced = 28 but 1/2 are consumed leaving 14
At 08:20 there are an extra 10 produced = 24 but 1/2 are consumed leaving 12
At 08:30 there are an extra 10 produced = 22 but 1/2 are consumed leaving 11
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4 0

2 years ago

2 1/8 × 1/3 estimate

horsena [70]

2 1/8 × 1/3
= 17/8 × 1/3
= 17/24

4 0

3 years ago

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vovikov84 [41]

Answer:

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Step-by-step explanation:

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3 years ago

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aliina [53]

Answer:

Step-by-step explanation:

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8 0

4 years ago

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$

Thus, the probability that a randomly selected woman shop less than 5 hours online is 0.9922.

8 0

4 years ago