Please help with these two questions
1 answer:
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Answer:
z = -1.645
Step-by-step explanation:
b) n =25, M = 4.01 , \mu = 4.22 , \sigma = 0.60 , \alpha= 0.05
The hypothesis are given by,
H0 : \mu\geq 4.22 v/s H1 : \mu < 4.22
The test statistic is given by,
Check attachment for the formula that should br here.
z = \frac{4.01- 4.22 }{0.60 /\sqrt{25}}
= -1.75
The critical value of z = -1.645
The calculated value z > The critical value of z
Hence we reject null hypothesis.
The healthy-weight students eat significantly fewer fatty, sugary snacks than the overall population.
The <span>given the piecewise function is :
</span>
![f(x) = \[ \begin{cases} 2x & x \ \textless \ 1 \\ 5 & x=1 \\ x^2 & x\ \textgreater \ 1 \end{cases} \]](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5C%5B%20%5Cbegin%7Bcases%7D%20%0A%20%20%20%20%20%202x%20%26%20x%20%5C%20%5Ctextless%20%5C%20%201%20%5C%5C%0A%20%20%20%20%20%205%20%26%20x%3D1%20%5C%5C%0A%20%20%20%20%20%20x%5E2%20%26%20x%5C%20%5Ctextgreater%20%5C%201%20%0A%20%20%20%5Cend%7Bcases%7D%0A%5C%5D)
To find f(5) ⇒ substitute with x = 5 in the function → x²
∴ f(5) = 5² = 25
To find f(2) ⇒ substitute with x = 5 in the function → x²
∴ f(2) = 2² = 4
To find f(-2) ⇒ substitute with x = 5 in the function → 2x
∴ f(-2) = 2 * (-2) = -4
To find f(1) ⇒ substitute with x = 1 in the function → 5
∴ f(1) = 5
================================
So, the statements which are true:<span>
</span><span></span>
</span>
To find f(5) ⇒ substitute with x = 5 in the function → x²
∴ f(5) = 5² = 25
To find f(2) ⇒ substitute with x = 5 in the function → x²
∴ f(2) = 2² = 4
To find f(-2) ⇒ substitute with x = 5 in the function → 2x
∴ f(-2) = 2 * (-2) = -4
To find f(1) ⇒ substitute with x = 1 in the function → 5
∴ f(1) = 5
================================
So, the statements which are true:<span>
Answer:
256
Step-by-step explanation:
A calculator works well for this.
_____
None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.
1 to any power is still 1, so the factors (1^n) can be ignored.
After you ignore all of the things that can be ignored, your problem simplifies to ...
(2^2)(2^-3)^-2
The rules of exponents applicable to this are ...
(a^b)^c = a^(b·c)
(a^b)(a^c) = a^(b+c)
Then your product simplifies to ...
(2^2)(2^((-3)(-2)) = (2^2)(2^6)
= 2^(2+6)
= 2^8 = 256
