Anybody answer me please with step by step explanation need complete detailed answer
2 answers:
The area of triangle is 55 square centimeters.
<h3>What is area?</h3>Area is the amount of space occupied by a two-dimensional figure.
<h3>What is the formula for the area of triangle?</h3>The formula for the area of triangle is
According to the given question.
We have a figure of a quadrilateral which is made up of four right angled tringles.
Now,
The area of a triangle with base 5cm and height 7 cm
=
square centimeters
And, the area of triangle with base 4cm and height 5cm
=
square centimeters
Since, the quadrilateral is made up of 2 right angled triangle with base 5cm and height 7cm and 2 right angled triangle with base 4cm and height 5cm.
Therefore, the area of quadrilateral or the given fiure
= 2( area of triangle with height 7cm and base 5cm + area of triangle with base 4cm and height 5cm )
= 2(17.5 + 10) square centimeter
= 2 × 27.5
= 55 square centimeters
Hence, the area of triangle is 55 square centimeters.
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Answer:
36 units
Step-by-step explanation:
Hope can help you ...I use the formula that to find its area...
Answer:
(a) 0.50928
(b) 0.857685.
Step-by-step explanation:
We are given that an engineer is going to redesign an ejection seat for an airplane. The new population of pilots has normally distributed weights with a mean of 155 lb and a standard deviation of 29.2 lb i.e.; = 160 lb and
= 27.5 lb
(A) We know that Z = ~ N(0,1)
Let X = randomly selected pilot
If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)
P(150 < X < 201) = P(X < 201) - P(X <= 150)
P(X < 201) = P( <
) = P(Z < 1.57) = 0.94179
P(X <= 150) = P( <
) = P(Z < -0.17) = 1 - P(Z < 0.17) = 1 - 0.56749
= 0.43251
Therefore, P(150 < X < 201) = 0.94179 - 0.43251 = 0.50928 .
(B) We know that for sampling mean distribution;
Z = ~ N(0,1)
If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);
P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)
P(X bar < 201) = P( <
) = P(Z < 9.84) = 1 - P(Z >= 9.84)
= 0.999995
P(X bar <= 150) = P( <
) = P(Z < -1.07) = 1 - P(Z < 1.07)
= 1 - 0.85769 = 0.14231
Therefore, P(150 < X bar < 210) = 0.999995 - 0.14231 = 0.857685.
C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.