Answer: the answer M< AJK = 90, M< BKJ = 90 and m< CLJ = 90. ALL 90
Step-by-step explanation: I just got it wrong and it told me the answer.
I am a number greater than 40,000 and less than 60,000:
40,000 < n < 60,000
This means that:
n = 10,000n₁ + 1,000n₂ + 100n₃ + 11n₄
And also:
4 ≤ n₁ < 6
0 ≤ n₂ ≤ 9
0 ≤ n₃ ≤ 9
0 ≤ n₄ ≤ 9
My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:
n₁ = 3*2n₄ - 1
n₁ = 6n₄ - 1
This means that:
n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄
n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄
n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃
<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:
n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:
n</span>₂ = 9 - n₃
<span>
Therefore:
9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:
n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:
</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6
n = 60,011n₄ - 10,000 + 3,000 + 600
n = 60,011n₄ - 6,400
Therefore:
0<n₄<2, so n₄=1.
If n₄=1:
n = 60,011 - 6,400
n = 53,611
Answer:
53,611
Answer:
Step-by-step explanation:
From the given information:
The null and the alternative hypothesis can be well written as:
Given that:
n = 200
x = 135
Alpha ∝ = 0.05 level of significance
Then;
⇒ 
= 200 × 0.6 × (1 -0.6)
= 200 × 0.6 × 0.4
= 48 ≥ 10
The sample proportion 
= 0.675
The test statistics 
Z = 2.165
The P-value = P(Z > 2.165)
= 1 - P(Z < 2.165)
From the z tables
= 1 - 0.9848
= 0.0152
Reject the null hypothesis since P-Value is lesser than alpha. ( i.e. 0.0152 < 0.05).
Thus, there is enough evidence to conclude that the value of the population proportion is greater than 0.6
Answer:
x = 12
Step-by-step explanation:
"Using the variable x for unknown number what is five more than the quotient of a number and 6 equals 7" translates into 5 + x/6 = 7.
To solve use inverse operations.
The mistake made by George is; D:George should have averaged the two differences instead of the two bounds.
<h3>How to Solve Successive Approximations?</h3>
In Mathematics, successive approximation can be defined as a classical method that is used in Calculus for solving integral equations or initial value problems.
In this question, George started the first iteration of successive approximation by using the lower and upper bounds of the graph. However, we can deduce that George made a mistake instep 5 because he should have used x = 3/2 as the new upper bound.
Read more about Successive Approximations at; brainly.com/question/25219621
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