Answer:
<em><u>Olá</u></em><em><u>,</u></em><em><u> </u></em><em><u>tudo</u></em><em><u> </u></em><em><u>bem</u></em><em><u>?</u></em><em><u> </u></em>
<em><u>Me</u></em><em><u> </u></em><em><u>chamo</u></em><em><u> </u></em><em><u>Ana</u></em><em><u> </u></em><em><u>Kesia</u></em><em><u>,</u></em><em><u> </u></em><em><u>muito</u></em><em><u> </u></em><em><u>prazer</u></em><em><u>.</u></em>
<em><u>Nn</u></em><em><u> </u></em><em><u>entendi</u></em><em><u> </u></em><em><u>sua</u></em><em><u> </u></em><em><u>pergunta</u></em><em><u>!</u></em><em><u> </u></em>
<em><u>Mais</u></em><em><u>,</u></em><em><u> </u></em><em><u>mesmo</u></em><em><u> </u></em><em><u>assim</u></em><em><u> </u></em><em><u>obrigada</u></em><em><u> </u></em><em><u>pelos</u></em><em><u> </u></em><em><u>pontos</u></em><em><u>.</u></em>
Answer:
Yes
Step-by-step explanation:
ΔMNL ≅ ΔQNL by ASA or AAS
by ASA
Proof:
∠ LNM = ∠LNQ =90
LN = LN {Common}
∠MLN = ∠QLN {LN bisects ∠ L}
By AAS
∠Q + ∠QLN + ∠LNQ = 180 {Angle sum property of triangle}
∠Q + 32 + 90 = 180
∠Q + 122 = 180
∠Q = 180 -122 =
∠Q = 58
∠Q = ∠M
∠MNL =∠QNL = 90
LN = LN {common side}
Answer: 0.0793
Step-by-step explanation:
Let the IQ of the educated adults be X then;
Assume X follows a normal distribution with mean 118 and standard deviation of 20.
This is a sampling question with sample size, n =200
To find the probability that the sample mean IQ is greater than 120:
P(X > 120) = 1 - P(X < 120)
Standardize the mean IQ using the sampling formula : Z = (X - μ) / σ/sqrt n
Where; X = sample mean IQ; μ =population mean IQ; σ = population standard deviation and n = sample size
Therefore, P(X>120) = 1 - P(Z < (120 - 118)/20/sqrt 200)
= 1 - P(Z< 1.41)
The P(Z<1.41) can then be obtained from the Z tables and the value is 0.9207
Thus; P(X< 120) = 1 - 0.9207
= 0.0793
Answer:
65
Step-by-step explanation:
First, let's consider the case where you have 1 apple in the basket. The possible number of baskets you can have with 1 apple is 11: the case where you have no oranges besides the apples, the case where you have 1 orange, the case where you have 2 orange, etc.
This line of reasoning can be followed by each possible amount of apples you can have: 0, 1, 2, 3, 4 and 5.
Keep in mind that in the case where you have no apples, you must exclude the case where you also have no oranges, because the problem tells you that the basket must have at least 1 fruit.
So, the answer would be:
x = 6*11 -1 = 65 possible baskets