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Yuliya22 [10]
3 years ago
13

Requie 2) 20 + (-45)

Mathematics
2 answers:
harkovskaia [24]3 years ago
6 0

Answer:

20 + (-45) = -25

daser333 [38]3 years ago
6 0
The answer is negative 25
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Question
AlekseyPX

Answer:

The population mean is 5 volts

Step-by-step explanation:

Output voltage is assumed to be normally distributed, with standard deviation 0.25 Volts

s=0.25

n = 8

H_0:\mu = 5\\H_a:\mu \neq 5

Sample mean = \bar{x}=4.85

Since n < 30 and population standard deviation is unknown

So, we will use t test

Formula : t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}

t = \frac{4.85-5}{\frac{0.25}{\sqrt{8}}}

t=-1.69

Refer the t table for p value

Degree of freedom = n-1 = 8-1 = 7

So,t_{(df,\alpha)}=t_{7,0.05}=2.365

P value >α

So, We are failed to reject null hypothesis

Hence The population mean is 5 volts

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3 years ago
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Answer:

All systems have 2 real solutions.

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8 0
3 years ago
Factor completely x^8- 16.
Nataly_w [17]

Answer:

Step-by-step explanation:


So in this example we'll be using the difference of squares which essentially states that: (a-b)(a+b)=a^2-b^2 or another way to think of it would be: a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}). So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:

x^8-16 = (x^4-4)(x^4+4)

Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:

x^4-4 = (x^2-2)(x^2+2)

So completely factored form is: (x^2-2)(x^2+4)(x^4+4)

I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because x^2+4 = x^2-(-4). and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: x^4+4 = x^4 - (-4) = (x^2-\sqrt{-4})(x^2+\sqrt{-4}) = (x^2-2i)(x^2+2i) just something that might be useful in some cases.

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