Answer:
D. Opposite angles are congruent
With option A, we can see that it is clearly wrong, in a parallelogram, only the opposite sides are congruent, if all of the sides are congruent then the shape is more likely to be a square or a rhombus.
With option B, only the opposites angles are congruent. If all four angles are congruent, that will be more likely to be true when we consider a square or a rectangle.
With option C, it is also wrong, unless the shape is a rhombus, square or rectangle.
That leaves us with the last option which is D.
Answer:
280 and 8
Step-by-step explanation:
the LCM is the lowest common multiple
the lowest multiple that can Divide 40 and 56 without a remainder = 280
the Hcf Is the highest common factor of 40 and 56
the highest number that can Divide both =8
Answer:
t ≥ 30
Step-by-step explanation:
Martha could exercise 30 min or more than 30 min each day.
Let t = her time exercising. Then
t ≥ 30
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Answer:
a) Binomial distribution, with n=20 and p=0.10.
b) P(x>1) = 0.6082
c) P(3≤X≤5) = 0.3118
d) E(X) = 2
e) σ=1.34
Step-by-step explanation:
a) As we have a constant "defective" rate for each unit, and we take a random sample of fixed size, the appropiate distribution to model this variable X is the binomial distribution.
The parameters of the binomial distribution for X are n=20 and p=0.10.
![X\sim B(0.10,20)](https://tex.z-dn.net/?f=X%5Csim%20B%280.10%2C20%29)
b) The probability of k defective surge protectors is calculated as:
![P(x=k) = \binom{n}{k} p^{k}q^{n-k}](https://tex.z-dn.net/?f=P%28x%3Dk%29%20%3D%20%5Cbinom%7Bn%7D%7Bk%7D%20p%5E%7Bk%7Dq%5E%7Bn-k%7D)
In this case, we want to know the probability that more than one unit is defective: P(x>1). This can be calculated as:
![P(x>1)=1-(P(0)+P(1))\\\\\\P(x=0) = \binom{20}{0} p^{0}q^{20}=1*1*0.1216=0.1216\\\\P(x=1) = \binom{20}{1} p^{1}q^{19}=20*0.1*0.1351=0.2702\\\\\\ P(x>1)=1-(0.1216+0.2702)=1-0.3918=0.6082](https://tex.z-dn.net/?f=P%28x%3E1%29%3D1-%28P%280%29%2BP%281%29%29%5C%5C%5C%5C%5C%5CP%28x%3D0%29%20%3D%20%5Cbinom%7B20%7D%7B0%7D%20p%5E%7B0%7Dq%5E%7B20%7D%3D1%2A1%2A0.1216%3D0.1216%5C%5C%5C%5CP%28x%3D1%29%20%3D%20%5Cbinom%7B20%7D%7B1%7D%20p%5E%7B1%7Dq%5E%7B19%7D%3D20%2A0.1%2A0.1351%3D0.2702%5C%5C%5C%5C%5C%5C%20P%28x%3E1%29%3D1-%280.1216%2B0.2702%29%3D1-0.3918%3D0.6082)
c) We have to calculate the probability that the number of defective surge protectors is between three and five: P(3≤X≤5).
![P(3\leq X\leq 5)=P(3)+P(4)+P(5)\\\\\\P(x=3) = \binom{20}{3} p^{3}q^{17}=1140*0.001*0.1668=0.1901\\\\P(x=4) = \binom{20}{4} p^{4}q^{16}=4845*0.0001*0.1853=0.0898\\\\P(x=5) = \binom{20}{5} p^{5}q^{15}=15504*0*0.2059=0.0319\\\\\\P(3\leq X\leq 5)=P(3)+P(4)+P(5)=0.1901+0.0898+0.0319=0.3118](https://tex.z-dn.net/?f=P%283%5Cleq%20X%5Cleq%205%29%3DP%283%29%2BP%284%29%2BP%285%29%5C%5C%5C%5C%5C%5CP%28x%3D3%29%20%3D%20%5Cbinom%7B20%7D%7B3%7D%20p%5E%7B3%7Dq%5E%7B17%7D%3D1140%2A0.001%2A0.1668%3D0.1901%5C%5C%5C%5CP%28x%3D4%29%20%3D%20%5Cbinom%7B20%7D%7B4%7D%20p%5E%7B4%7Dq%5E%7B16%7D%3D4845%2A0.0001%2A0.1853%3D0.0898%5C%5C%5C%5CP%28x%3D5%29%20%3D%20%5Cbinom%7B20%7D%7B5%7D%20p%5E%7B5%7Dq%5E%7B15%7D%3D15504%2A0%2A0.2059%3D0.0319%5C%5C%5C%5C%5C%5CP%283%5Cleq%20X%5Cleq%205%29%3DP%283%29%2BP%284%29%2BP%285%29%3D0.1901%2B0.0898%2B0.0319%3D0.3118)
d) The expected number of defective surge protectors can be calculated from the mean of the binomial distribution:
![E(X)=\mu_B=np=20*0.10=2](https://tex.z-dn.net/?f=E%28X%29%3D%5Cmu_B%3Dnp%3D20%2A0.10%3D2)
e) The standard deviation of this binomial distribution is:
![\sigma=\sqrt{np(1-p)}=\sqrt{20*0.1*0.9}=\sqrt{1.8}=1.34](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B20%2A0.1%2A0.9%7D%3D%5Csqrt%7B1.8%7D%3D1.34)