2/3 of a class are girls. A) what is the ratio girls to boys in the class? B) what is the ratio boys to girls in the class

The first quartile of a data set is 23, the median is 30,the third quartile is 33, and an outlier is 50. Which of these data val

Len [333]

Answer:50

Step-by-step explanation:

6 0

3 years ago

What is the additive inverse of 7/2 ? A) −2/7 B) −7/2 C) 2/7 D) 7/2

snow_lady [41]

The additive inverse of 7/2 would be -7/2 (option B) Since its. opposite !

A)is the additive inverse of C) ; C) is the additive inverse of A) (not your answer)

B. is the additive inverse for this problem (7/2) so thats how we found that this is your answer (answer)

D is the same thing for the problem (7/2) (not your answer)

7/2 isnt the additive inverse for 7/2 because they are the SAME not the opposite.

+ = - (Positive = negative)

- = + (negative = positive)

6 0

3 years ago

How would i solve for x and y in 3x+2y=17 and 2x-y=2 by using substitution

maxonik [38]

I’ll do an example problem, and I challenge you to do this on your own!
4x+6y=23
7y-8x=5
Solving for y in 4x+6y=23, we can separate the y by subtracting both sides by 4x (addition property of equality), resulting in 6y=23-4x. To make the y separate from everything else, we divide by 6, resulting in (23-4x)/6=y. To solve for x, we can do something similar - subtract 6y from both sides to get 23-6y=4x. Next, divide both sides by 4 to get (23-6y)/4=x.

Since we know that (23-4x)/6=y, we can plug that into 7y-8x=5, resulting in
7*(23-4x)/6-8x=5
= (161-28x)/6-8x
Multiplying both sides by 6, we get 161-28x-48x=30
= 161-76x
Subtracting 161 from both sides, we get -131=-76x. Next, we can divide both sides by -76 to separate the x and get x=131/76. Plugging that into 4x+6y=23, we get 4(131/76)+6y=23. Subtracting 4(131/76) from both sides, we get
6y=23-524/76. Lastly, we can divide both sides by 6 to get y=(23-524/76)/6

Good luck, and feel free to ask any questions!

4 0

3 years ago

In the parallelogram, M

alexira [117]

Answer:

112°

Step-by-step explanation:

Given parallelogram JKLM with:

m∠KLO = 53°
m∠MLO = 59°

Find:

m∠KJM

We know opposite angles of parallelogram are congruent:

m∠KJM = m∠KLM

Use angle addition postulate:

m∠KLM = m∠KLO + m∠MLO
m∠KLM = 53° + 59°
m∠KLM = 112°

5 0

3 years ago

Simplify the following Questions. (IMPORTANT NOTE!!! #'s 2 AND 7 MUST BE IN SCIENTIFIC NOTATION!!!!!) 1.) (x)²(2xy³)^5 2.) (4x10

poizon [28]

Answer:

Step-by-step explanation:

Hint :

$(a*b)^{m}= a^{m}b^{m}\\\\\\(a^{m})^{n}=a^{mn}\\\\\\a^{m}*a^{n}=a^{m+n}\\$

$1)x^{2}(2xy^{3})^{5}=x^{2}*2^{5}*x^{5}*(y^{3})^{5}\\\\\\=x^{2}*2^{5}*x^{5}*y^{3*5}\\\\=x^{2}*2^{5}*x^{5}*y^{15}\\\\=2^{5}*x^{2+5}*y^{15}\\\\=2^{5}x^{7}y^{15}$

$2) (4x10^{8})^{2}=4^{2}*x^{2}*(10^{8})^{2}\\\\=16*x^{2}*10^{8*2}\\\\=1.6*10*x^{2}*10^{16}\\\\=1.6x^{2}*10^{16+1}\\\\=1.6x^{2}10^{17}$

$3)(-h^{4})^{5}=(-h)^{4*5}=-h^{20}\\\\\4)(3xy^{3})^{2}(xy)^{6}=3^{2}x^{2}(y^{3})^{2}x^{6}y^{6}\\\\=9x^{2}y^{3*2}x^{6}y^{6}\\\\\\=9x^{2}y^{6}x^{6}y^{6}\\\\=9x^{2+6}y^{6+6}\\\\=9x^{8}y{12}\\\\\\$

$5) (p^{9})^{-2}=p^{9*-2}=p^{-18}\\\\\\6)(5k^{2})^{3}=5^{3}(k^{2})^{3}=625k^{6}\\\\7)(7x10^{5})^{2}=7^{2}x^{2}(10^{5})^{2}\\\\=49x^{2}10^{5*2}\\\\=4.9*10^{1}x^{2}10^{10}\\\\=4.9x^{2}10^{10+1}\\\\=4.9x^{2}10^{11}$

$8)x^{3}(-x^{3}y)^{2}=x^{3}*(-x^3})^{2}y^{2}\\\\\\=x^{3}*(-1)^{2}*x^{2*3}y^{2}\\\\=x^{3}*1*x^{6}y^{2}=x^{3+6}y^{2}\\\\=x^{9}y^{2}\\\\\\9)w^{5}(w^{2})^{-4}=w^{5}w^{2*-4}\\\\=w^{5}w^{-8}\\\\=w^{5-8}=w^{-3}\\\\=\frac{1}{w^{3}}\\\\\\10) (2x^{5})^{4}=2^{4}x^{5*4}\\\\=2^{4}x^{20}\\\\=16x^{20}$

5 0

3 years ago