Answer:
Ok
Step-by-step explanation:
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Answer:
The mean of the sampling distribution of p is 0.75
The standard deviation of the sampling distribution of p is 0.0274
Step-by-step explanation:
According to the given data 75% of the sampled households watch sports on television at least once a month therefore the mean of the sampling distribution of p is 0.75.
In order to calculate the standard deviation of the sampling distribution of p we would have to use the following formula as follows:
Standard deviation = √ [ p ( 1 - p) / n ]
= √[ 0.75 * ( 1 - 0.75) / 250]
= 0.0274
The standard deviation of the sampling distribution of p is 0.0274.
Let us assume the first odd number = x
Then the second consecutive odd number = x + 2
The third consecutive odd number = x + 4
The addition of the three consecutive odd numbers = 75
Then we can write the equation as
x + x + 2 + x + 4 = 75
3x + 6 = 75
3x = 75 - 6
3x = 69
x = 69/3
= 23
Then
The greatest odd number in the question = X + 4
= 23 + 4
= 27
So the greatest odd number in the question is 27.
We know that
<span>The measure of the external angle is the semi difference of the arcs that it covers.
so
see the picture attached to better understand the problem
m AB=128</span>°--------> by central angle
<span>m ACB=360</span>°-128°-----> m ACB=232°<span>
</span>∠x=(1/2)*[m ACB-m AB]<span>
</span>∠x=(1/2)*[232-128]-----> ∠x=(1/2)*104°-----> ∠x=52°
<span>
the answer is</span>
∠ x is 52°<span>
</span>
According to the Pythagorean triple, the side lengths that will form a right triangle are:
A. 11, 13, √290
B. 7, 24, 25
<h3>What is the Pythagorean Triple?</h3>
The Pythagorean triple are three set of sides that can form a right triangle if the square of the longest side equals the sum of the squares of the other two smaller sides.
11² + 13² = (√290)²
290 = 290 [correct]
7² + 24² = 25²
625 = 625 [correct]
2² + 5² = 7²
29 ≠ 49 [correct]
Therefore, the side lengths that will form a right triangle are:
A. 11, 13, √290
B. 7, 24, 25
Learn more about the Pythagorean triple on:
brainly.com/question/11994763
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