Answer:
B) 5
Step-by-step explanation:
f(-2) = (-2)² + 5•(-2) +11
f(-2) = 4 -10 +11
f(-2) = 5
Answer: See attachment below (graphing)
Step-by-step explanation:
<u>INFORMATION:</u>
The slope-intercept is when y=mx+b
mx=the slope
b=y-intercept
<u />
<u>REPHRASE EQUATION:</u>
y=mx+b
y=-3-x
y=(-1)x-3
<u>CONCLUSION:</u>
Slope=-1
Y-intercept=-3
Hope this helps!! :)
Please let me know if you have any questions
The best way to compare fractions would be to make them have like
denominators. We first , in this case, need to convert from decimal to
fraction.
Converting decimals to fractions first requires an
understanding of the decimal places that fall after the decimal. One
place after the decimal is the tenths place. If you have a decimal that
ends at one place after the decimal (or in the tenths place) it can be
written as the number after the decimal in the top of the fraction and
ten (tenths place) in the denominator. ex. .5 ends one place after
the decimal and can be written as 5/10...(read as five tenths).
If a decimal ends at two places after the decimal...(ex. .75)...it
ends in the hundredths place, can be written as that number in the
numerator and 100 in the denominator....(ex 75/100) and is read as
seventy-five hundredths.
one place after the decimal is tenths (over 10), two places is
hundredths (over 100), three places is thousandths (over 1000) , four
places ten-thousandths (over 10000) and so on.
Because each decimal in your problem has a different amount of
decimal places, it makes for different denominators. But, We can add a
zero to the end of a decimal without changing it's value; if we add a
zero to the end of .5 and make it .50 , we then can write it as 50/100
and would now have like denominators.
if .5 = .50 = 50/100 and .75 = 75/100
we now have the question what fractions can fall between 50/100 and 75/100.
That would be fractions such as 51/100, 52/100, 53/100.......74/100.
Answer:
The correct option is;
B. I only
Step-by-step explanation:
The central limit theorem states that when adequately large random samples are taken from a population that has a mean, μ, and standard deviation, σ, the distributions of the sample means of individual samples taken will be more or less (approximately) normally distributed
The central limit theorem is correct for both normal and skewed populations such that estimates can be made about the population mean based on available sample mean information.
I can’t see your image, can you select another one by any chance ?
