The following data show the scores Gina obtained on 12 IQ tests:

Mathematics

1 answer:

krek1111 [17]3 years ago

8 0

Answer:
The Upper Quartile is incorrect

Explanation:
Hope it helps :)

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Using the completing-the-square method, rewrite f(x) = x2 - 6x + 2 in vertex form.<br>

sp2606 [1]

Answer:

$\large\boxed{f(x)+(x-3)^2-7}$

Step-by-step explanation:

$\text{The vertex form of equation}\ y=ax^2+bx+c:\\\\y=a(x-h)^2+k\\\\\text{We have the equation:}\\\\f(x)=x^2-6x+2=x^2-2(x)(3)+1=x^2-2(x)(3)+3^2-3^2+2\\\\\text{Use}\ (a-b)^2=a^2-2ab+b^2\to x^2-2(x)(3)+3^2=(x-3)^2\\\\f(x)=(x-3)^2-9+2=(x-3)^2-7$

8 0

3 years ago

Patrick buys 12 bunches of bananas for $9 what will he pay for 8 bunches of bananas

larisa [96]

$12b.......9\$ \\ 8b.........x \\\\ x=\frac{8*9}{12}=\frac{72}{12} \\\\ \boxed{x=5\$}$

5 0

3 years ago

In triangle abc, m of acb = 90, cd is perpendicular to ab , m of acd is 60. and bd is 5 cm. find ad

weeeeeb [17]

Let us draw a picture to make the things more clear.

Attached is the image.

We have been given that

$\angle acd = 60 ^{\circ}$

Therefore, we have

$\angle dcb =90- 60= 30 ^{\circ}$

Now, in triangle bcd, we have

$\tan30 = \frac{5}{cd}\\
\\
\frac{1}{\sqrt 3}=\frac{5}{cd}\\
\\
cd=5\sqrt 3$

Now, in triangle acb, we have

$tan 60 = \frac{ad}{5\sqrt3} \\
\\
\sqrt 3= \frac{ad}{5\sqrt3}\\
\\
ad= 5\sqrt3 \times \sqrt 3\\
\\
ad= 5\times 3\\
\\
ad=15$

Thus, ad is 15 cm.

4 0

3 years ago

What is the average rate of change for 5/6 and 15/14

andreev551 [17]

Answer:

Average rate of change is <u>0.80.</u>

Step-by-step explanation:

Given:

The two points given are (5, 6) and (15, 14).

Average rate of change is the ratio of the overall change in 'y' and overall change in 'x'. If the overall change in 'y' is positive with 'x', then average rate of change is also positive and vice-versa.

The average rate of change for two points $(x_1,y_1)\ and\ (x_2,y_2)$ is given as: