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Marina86 [1]
3 years ago
12

50 POINTS!!! Type the correct answer in the box. Use numerals instead of words.

Mathematics
2 answers:
ira [324]3 years ago
7 0

Answer:

0.7835140

Step-by-step explanation:

Given that:

Population mean = 62.95

Population standard deviation = 5.65

Sample size = 52

The standard error of the sample mean :

Standard deviation / sqrt(sample size)

5.65 / sqrt(52)

= 0.7835140

djyliett [7]3 years ago
6 0

Answer:

0.78

Step-by-step explanation:

standard deviation is 5.65

sample size is 52

\frac{5.65}{\sqrt{52} } = 0.789514 rounded to the nearest hundredth is 0.78

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- 3.3 - ? = 2 1/2 = what value make the equation true ?
Zigmanuir [339]
? = 5.8

-3.3 - 5.8 = 2.5 or 2 1/2
4 0
3 years ago
If f (7) = 22, then<br> ff-1(22)) = [?]
lozanna [386]

Answer:

f(f^{-1} (22)) equals 22.

Step-by-step explanation:

For every input in a function, there will be one corresponding output. When taking the output of the function and inserting it as the input to the inverse function of the original, the output will be equivalent to the exact input of the original function. In this case, because f(7)=22, that means f^{-1} (22) will equal 7. Using this knowledge, you can find the value of f(f^{-1} (22)), which will be f(f^{-1} (22)) = f(7)=22, so f(f^{-1} (22)) equals 22.

8 0
3 years ago
Which expressions are equivalent to 12x + 10 + 4y?<br><br>OPTIONS are in the picture​
yan [13]

Answer:

The answer is D.

Step-by-step explanation:

2(6x+5+2y)

Use the distributive property

2(6x)+2(5)+2(2y)

Simplify

12x+10+4y

Hope this helps ^-^

3 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=x%20%2B%20y%20%2B%20%20%7Bx%7D%5E%7B2%7D%20%20-%20%20%7By%7D%5E%7B2%7D%20" id="TexFormula1" ti
Alexus [3.1K]

Answer:

(x + y)(1 + x - y)

Step-by-step explanation:

x + y + x² - y²

(x + y) + (x - y)(x + y)

[ Using a²-b² = (a-b)(a+b) ]

Taking (x + y) common

(x + y)(1 + x - y)

5 0
4 years ago
I wasted 10 points for this, but if u seen this link
Firdavs [7]
Ok thank you for letting me know
3 0
3 years ago
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