<span>To link the displayed Wright flyer in the museum with the
actual plane, we have to calculate for the relation of the corresponding
parameter. In this problem, for both the display and the actual plane, we are
given with the length. Computing for the ratio gives us,</span>
<span>
ratio = length of the model / length of the actual plane
Length of the model = 35 cm</span>
<span>
We need to calculate for the length of the model in ft.</span>
<span>
length of the model = (35 cm)(1 in/2.54 cm)(1 ft/12 in)
length of the model = 1.148 ft
</span>
<span>
</span>
<span>Going back to the calculation for the ratio,
ratio = (1.148 ft)/21 ft
ratio = 0.055
Therefore, the measurements used in the model is equal to 0.055 times the
actual dimensions.
Error may occur because of the number of significant figures measured for
rounding up or down of the answers after each calculation. </span>
Answer:
Scatter plots resembles line graphs in that they are plotted on the x and y axis. The are used because they show how much one variable is affected by another.
Scatter diagrams are important since in statistics they can show the extent of correlation, if any, between the values of observed quantities or a behavior.
Answer:
4 shelves
Step-by-step explanation:
if there is 8 shelves that display 1 car then that means there is 8 cars total.
so if you need a total of 12 cars then you would do (total of cars needed - total of cars you have).
(12-8) which would equal 4.
you need 4 cars left but on each shelf going forward needs to have 2 cars on it.
now since you have 4 cars you would divide that by the amount of cars on each shelf going forward.
4/2) which equals 2.
you need 2 more shelfs for the 4 cars needed.
Answer:
∠N and ∠Y
Step-by-step explanation:
When two triangles are congruent then there corresponding parts ( angle or sides ) are also congruent or equal, ( CPCTC )
In triangles MNO and XYZ,
Side MN, NO and OM are corresponding to XY, YZ and ZX respectively,
And, angles M, N and O are corresponding to angles X, Y and Z respectively,
Thus, if ΔMNO ≅ ΔXYZ
By CPCTC,
MN ≅ XY
NO ≅ YZ
OM ≅ ZX
Also, ∠M ≅ ∠X, ∠N ≅ ∠Y and ∠O ≅ ∠Z
Hence, THIRD OPTION is correct.
Answer:
a) Calculate the probability that at least one of them suffers from arachnophobia.
x = number of students suffering from arachnophobia
= P(x ≥ 1)
= 1 - P(x = 0)
= 1 - [0.05⁰ x (1 - 0.05)¹¹⁻⁰
]
= 1 - (0.95)¹¹
= 0.4311999 = 0.4312
b) Calculate the probability that exactly 2 of them suffer from arachnophobia? 0.08666
= P(x = 2)
= (¹¹₂) x (0.05)² x (0.95)⁹
where ¹¹₂ = 11! / (2!9!) = (11 x 10) / (2 x 1) = 55
= 55 x 0.0025 x 0.630249409 = 0.086659293 = 0.0867
c) Calculate the probability that at most 1 of them suffers from arachnophobia?
P(x ≤ 1)
= P(x = 0) + P(x = 1)
= [(¹¹₀) x 0.05⁰ x 0.95¹¹] + [(¹¹₁) x 0.05¹ x 0.95¹⁰]
= (1 x 1 x 0.5688) + (11 x 0.05 x 0.598736939) = 0.5688 + 0.3293 = 0.8981
