How much does 9cm equal to inches

Mathematics

2 answers:

My name is Ann [436]3 years ago

6 0

Answer: 3.5 inches

When converted, 9cm is equal to 3.5in.

xxMikexx [17]3 years ago

4 0

Answer:

Exactly 3.54331

Step-by-step explanation:

Since 1 cm = 0,39 inches, just multiply 0,39 by 9 and there is your answer.

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Consider the following hypothesis test: H 0: 20 H a: < 20 A sample of 60 provided a sample mean of 19.5. The population stand

viva [34]

Answer:

We reject the null hypothesis and accept the alternate hypothesis. Thus, it be concluded that the population mean is less than 20.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 60

Sample mean, $\bar{x}$ = 19.5

Sample size, n = 60

Alpha, α = 0.05

Population standard deviation, σ = 1.8

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 20\\H_A: \mu < 20$

We use One-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{19.5 - 20}{\frac{1.8}{\sqrt{60}} } = -2.151$

Now, $z_{critical} \text{ at 0.05 level of significance } = -1.64$

Since,

$z_{stat} < z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis. Thus, it be concluded that the population mean is less than 20.

5 0

3 years ago

Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located

arsen [322]

Check the picture below.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$