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3 years ago

You tell me can A triangle can be formed with side lengths 4 in, 5 in, and 8 in.

1 answer:

5 0

Yes because the adding 4 and 8 would give you the max 12. Then to get the minimum you would subtract 8 and 4 getting 4. The numbers have to fall within that spectrum. They cannot be any more, less, or the exact numbers given.

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The time needed to complete a final examination in a particular college course is normally distributed with a mean of 77 minutes

Komok [63]

Answer:

a. The probability of completing the exam in one hour or less is 0.0783

b. The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. The number of students will be unable to complete the exam in the allotted time is 8

Step-by-step explanation:

a. According to the given we have the following:

The time for completing the final exam in a particular college is distributed normally with mean (μ) is 77 minutes and standard deviation (σ) is 12 minutes

Hence, For X = 60, the Z- scores is obtained as follows:

Z= X−μ /σ

Z=60−77 /12

Z=−1.4167

Using the standard normal table, the probability P(Z≤−1.4167) is approximately 0.0783.

P(Z≤−1.4167)=0.0783

Therefore, The probability of completing the exam in one hour or less is 0.0783.

b. In this case For X = 75, the Z- scores is obtained as follows:

Z= X−μ /σ

Z=75−77 /12

Z=−0.1667

Using the standard normal table, the probability P(Z≤−0.1667) is approximately 0.4338.

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is obtained as follows:

P(60<X<75)=P(Z≤−0.1667)−P(Z≤−1.4167)

=0.4338−0.0783

=0.3555

Therefore, The probability that student will complete the exam in more than 60 minutes but less than 75 minutes is 0.3555

c. In order to compute how many students you expect will be unable to complete the exam in the allotted time we have to first compute the Z−score of the critical value (X=90) as follows:

Z= X−μ /σ

Z=90−77 /12

Z=1.0833

UsING the standard normal table, the probability P(Z≤1.0833) is approximately 0.8599.

Therefore P(Z>1.0833)=1−P(Z≤1.0833)

=1−0.8599

=0.1401

Therefore, The number of students will be unable to complete the exam in the allotted time is= 60×0.1401=8.406

The number of students will be unable to complete the exam in the allotted time is 8

6 0

3 years ago

Find the area under the curve y = 27/x3 from x = 1 to x = t. Evaluate the area under this curve for t = 10, t = 100, and t = 100

Darina [25.2K]

Answer:

Step-by-step explanation:

given is a function as

$y=\frac{27}{x^3}$

We are to find the area form x=1 to x=t

The curve from x=1 lies in the I quadratnt.

So area above x axis is to be calculated

Area = $\int\limits^t_1 {\frac{27}{x^3} } \, dx \\=\frac{-27}{2x^2} \\= \frac{-27}{2t^2}-\frac{-27}{2}\\=\frac{27}{2}(1-\frac{1}{t^2} )$

a) When t =10,

area = $\frac{27}{2} (1-\frac{1}{10^2} )\\= 13.365$

b) t=100

area = $\frac{27}{2} (1-\frac{1}{100^2} )\\= 13.49865$

c) t=10000

area = $\frac{27}{2} (1-\frac{1}{10000^2} )\\= 13.5$

6 0

3 years ago

The distance d of a particle moving in a straight line is given by d(t) = 2t3 + 5t – 2, where t is given in seconds and d is mea

Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

Given the distance, d(t) of a particle moving in a straight line at any time t is:

$d(t) = 2t^3 + 5t - 2, $ where t is given in seconds and d is measured in meters.$

To find an expression for the instantaneous velocity v(t) of the particle at any given point in time, we take the derivative of d(t).

$v(t)=\dfrac{d}{dt}\\\\v(t) =\dfrac{d}{dt}(2t^3 + 5t - 2) =3(2)t^{3-1}+5t^{1-1}\\\\v(t)=6t^2+5$

The correct option is C.

6 0

3 years ago

A 30-minute TV program consists of three commercials, each 2 1/2 minutes long, and four equal-length entertainment segments. How

Alexus [3.1K]

for one show it will be 23.1 minutes if you multiply 2.30 minutes by 3 and get 6.9 you subtract 6.9 from the 30 minutes and you should 23.1 minutes

6 0

3 years ago

Read 2 more answers

A particular paper included the accompanying data on the tar level of cigarettes smoked for a sample of male smokers who subsequ

Dafna1 [17]

Answer:

Check the explanation

Step-by-step explanation:

$H_0$ :proportion of male smoker lung deaths is same for the four given tar level categories.

$H_1$ :proportion of male smoker lung deaths is not the same for the four given tar level categories.

Expected frequency=1177/4=294.25

Tar level Observed Freq.(O) Expected Freq.(E) (O-E)^2/E

0-7 107 294.25 120.435

8-14 375 294.25 5.643

15-21 553 294.25 227.533

>=22 183 294.25 42.061

Total= 1177 1177 395.673

Total chi square score=395.673

df=4-1=3

p-value=CHIDIST(395.673,3)<0.001

p-value<0.001,Reject null hypothesis.

There is sufficient evidence that the proportion of male smoker lung deaths is not the same for the four given tar level categories.

5 0

3 years ago