How do you solve 5 3/7 ÷ 5/11

Sphinxa [80]

Rational numbers are whole integers w no fractions. bc the square root of 72 is not a whole number, it is irrational

4 0

3 years ago

Which of the following conditions must be met in order to make a statistical inference about a population based on a sample

klasskru [66]

Answer:

For this case if we want to conclude that the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:

$\bar X \sim (\mu, \frac{\sigma}{\sqrt{n}})$

For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:

n>= 30

Step-by-step explanation:

For this case if we want to conclude that the sample does not come from a normally distributed population we need to satisfy the condition that the sample size would be large enough in order to use the central limit theoream and approximate the sample mean with the following distribution:

$\bar X \sim (\mu, \frac{\sigma}{\sqrt{n}})$

For this case the condition required in order to consider a sample size large is that n>30, then the best solution would be:

n>= 30

7 0

3 years ago

The Vertex is the part of a quadratic equation where the graph changes_________?

Georgia [21]

I think it slopeeeeeeeeeeeeeeeeeeee

8 0

3 years ago

Read 2 more answers

How to solve this :') please help

blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

$\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\$

Use $\cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1$

so

$\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}$

7 0

2 years ago

What type of graph is this?

murzikaleks [220]

Answer:

The graph of a quadratic equation, or a parabola, looks like a U, an upside down U, a C, or a backwards C. We can use the following rules to determine what the graph of a given quadratic equation looks like. If y = ax2 + bx + c, and a is positive, then the graph of the equation is the shape of a U.

Step-by-step explanation:

bc

5 0

3 years ago