Question

Again can someone plz help me before my head explodes

Answer 1

7.)

Mean: 4

Median: 3

Mode: 2

The rest I can't say but you can use this to help you: m a t h w a y . c o m / A l g e b r a

Answer 2

easy peasy,

1.

$\frac{\sqrt{3c^2}}{\sqrt{27}} \\ \text{since both terms inside the square root are positive, you can take it in root as a whole. (if there's a negative sign inside root, you can't do that)} \\ \sqrt{\frac{3c^2}{27}} \\\text{cancel the common factor 3} \\ \sqrt{\frac{c^2}{9}} \\ \text{now, simplify the exponents c^2 and 9=3^2 } \\ \text{property used:} \\ (a^m)^n=a^{m\times n} \\ \frac{c^{2\times\frac{1}{2}}}{3^{2\times\frac{1}{2}}}=\frac{c}{3}$

Note : i took the whole fraction in square root because they had the same power i.e. ½, had it been a square root and cube root, you could haven't done that.

2ⁿᵈ Method

or you can simplify the exponents from the beginning,

$\frac{\sqrt{3c^2}}{\sqrt{27}} \\ = \frac{3^{\frac{1}{2}}\times c^{2\times\frac{1}{2}}}{3^{3\times\frac{1}{2}}} \\\text{now, since the base is same, use the property: } \\ \frac{a^m}{b^n}=a^{m-n} \\=3^{\frac{1}{2}-\frac{3}{2}} \times c^1\\ = 3^{-1} \ c \\\text{property:} \\ a^{-m}=\frac{1}{a^m}\\ =\frac{c}{3}$

basically, it's all about properties of exponents, both methods are same.

2.

$\sqrt{27} \sqrt{3} \\ \text{similarly, you can either multiply the terms in root first,} \\ = \sqrt{27 \times 3} = \sqrt{81} \\ \text{now, 81= 3^4 (if you don't know, you can prime factorize 81)} \\ =\sqrt{3^4} \\ \text{ again multiply the exponents by the property:} \\ (a^m)^n=a^{m\times n} \\ =3^{4\times \frac{1}{2}} \\ = 3^2 \\ = 9$

2ⁿᵈ Method

note, $27=3^3$

simplify the exponents first, by the property:

$(a^m)^n=a^{m\times n} \\ = 3^{3\times \frac{1}{2}} \times 3^{\frac{1}{2}} \\ \text{now, since the base is same, use the property:} \\ a^m \times a^n = a^{m+n} \\ =3^{\frac{3}{2}+\frac{1}{2}} \\ =3^2 \\=9$

3.

$\sqrt{75}-4\sqrt{75} \\ \text{first, subtract the terms with common radical terms which is \sqrt{75} } \\ =-3\sqrt{75} \\ \text{ now} 75= 25 \times 3 \text{or} 75=5^2\times3 \\ \text{so,} \\ =-3\times (5^{2\times \frac{1}{2}}\times3^{\frac{1}{2}}) \\ = -15\sqrt3$

4.

$(\sqrt3+\sqrt2)^2\\ \text{note that } \sqrt3+\sqrt2 \text{ is already in the simplest form, you can't do anything about it to simplify it more } \\ \text{ so, expand the square by} \\ (a+b)^2=a^2+b^2+2ab \\ (\sqrt3+\sqrt2)^2 \\ = (\sqrt3)^2+(\sqrt2)^2 +2\sqrt3\sqrt2 \\ = 3+5+2\sqrt{3\times2} \\= 5+2\sqrt6$

5.

$\text{domain of } y=\sqrt{x+5} \\ \text{for a real valued function,} \\ \text{even radical signs ( \sqrt[2]{5} or simply \sqrt5 , \sqrt[6]{3} } \\ \text{ or fractional power with even denominators \frac{3}{10} etc.)} \\ \text{must NEVER have negative values, (do you know why?)} \\ \text{so,} \\ (x+5)\ge 0 \\ x\ge -5 \\ \text{hence the domain is} [-5,\infty)$

can't plot the graph here, but if you know the graph of $\sqrtx$ the vertex just shifts at $(-5,0) \text{ for } \sqrt{x+5}$

here's the image from Desmos

6.

I don't want to, actually I forgot it. but even if I knew, I wouldn't have done that. I dislike statistics.

7. I think mean best describes the data among Mean, median and Mode. I can't explain.

Mean is average value, $\frac{\text{sum of observations}}{\text{number of observations}}$

mode is the data value which occurs the most in data or alternatively, whose frequency is the highest

Median is the middle value of an ordered data (in ascending order)

for odd number of observations, median is the

$\left(\frac{n+1}{2}\right)^{\text{th}}$ term

and for even number of observations median is the mean of $\left(\frac{n}{2}\right)^{\text{th}}$ term and $\left(\frac{n}{2}+1\right)^{\text{th}}$ term,

that is, add both terms and divide by $2$,

here $n$ is the total number of observations.

Clearly, Mode is $2$ here,

arrange the data in ascending order,

$2 , 2, 2, 2, 3, 3, 5, 5, 7, 9$

terms are even so, median =

$\dfrac{\left(\frac{n}{2}\right)^{\text{th}}\text{term} +\left(\frac{n}{2}+1\right)^{\text{th}} \text{term}}{2}\\ = \frac{5^{\text{th}}\text{term}+6^{\text{th}}\text{term}}{2} \\\ = \frac{3+3}{2} \\ = 3$

for mean,

$\frac{2+2+2+2+3+3+5+5+7+9}{10} \\ =\frac{40}{10} \\ = 4$

8.

8.given that you ran $2 , 4 \text{ and } 5$ miles

miles let, $x \text{ be the miles ran of} 4^{text{th}}$ day,

you want the mean miles ran of the these 4 days to be 4

so,

$\frac{2+4+5+x}{4}=4$

$11+x=16$

$x=5$

hence, you should run 5 miles to make the mean of 4 miles per day for these 4 days.

9. I don't know what box-and-whisker is

10. I don't know what box-and-whisker is

hope this is clear Enough.