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1 answer:
Answer: 23 ft (approximately)
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Explanation:
Draw out a diagram as I've done so below. The unknown side x is opposite the reference angle 68. The hypotenuse (ladder length) is 25.
We'll use the sine rule
sin(angle) = opposite/hypotenuse
sin(68) = x/25
25*sin(68) = x
x = 25*sin(68)
x = 23.1795963641697
x = 23
The ladder reaches approximately 23 feet up on the wall.
I'm rounding to the nearest whole number because the other values given to you were also whole numbers. If your teacher says to round otherwise, then follow those instructions instead.
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Answer:
Step-by-step explanation:
The area of a triangle is given by the formula:
● A = (b×h)/2
b is the base and h is the heigth.
■■■■■■■■■■■■■■■■■■■■■■■■■■
Draw the heigth. This gives you two small right triangles.
Let's focus on the right one containing 45°.
● sin 45° = h/8
● h = sin 45° × 8 = 4√(2)
■■■■■■■■■■■■■■■■■■■■■■■■■■
Replace h by its value in the area formula. The base is 17
● A = (17× 4√(2))/ 2
● A = 34√(2)
● A = 48.08
Round to the nearest unit
● A = 48
Answer:
a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548
b)
- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504
- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177
- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281
Step-by-step explanation:
Given the data in the question;
proportion p = 0.6
sample size n = 10
binomial distribution
let x rep number of orders for raw materials arriving late in the sample.
(a) probability of committing a type I error if the true proportion is p = 0.6;
∝ = P( type I error )
= P( reject null hypothesis when p = 0.6 )
= ³∑ b( x, n, p )
= ³∑ b( x, 10, 0.6 )
= ³∑
∝ = 0.0548
Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548
b)
the probability of committing a type II error for the alternative hypotheses p = 0.3
β = P( type II error )
= P( accept the null hypothesis when p = 0.3 )
= ¹⁰∑ b( x, n, p )
= ¹⁰∑ b( x, 10, 0.3 )
= ¹⁰∑
= 1 - ³∑
= 1 - 0.6496
= 0.3504
Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504
the probability of committing a type II error for the alternative hypotheses p = 0.4
β = P( type II error )
= P( accept the null hypothesis when p = 0.4 )
= ¹⁰∑ b( x, n, p )
= ¹⁰∑ b( x, 10, 0.4 )
= ¹⁰∑
= 1 - ³∑
= 1 - 0.3823
= 0.6177
Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177
the probability of committing a type II error for the alternative hypotheses p = 0.5
β = P( type II error )
= P( accept the null hypothesis when p = 0.5 )
= ¹⁰∑ b( x, n, p )
= ¹⁰∑ b( x, 10, 0.5 )
= ¹⁰∑
= 1 - ³∑
= 1 - 0.1719
= 0.8281
Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281