Ask question

Ask question

All categories

3 years ago

6

Sue took a test in both biology and math last week. The biology test had a mean of 70 and a standard deviation of 7 whereas the

math test had a mean of 75 and a standard deviation of 10. Sue scored a 76 on the biology test and a 76 on the math test. On which test did she do better in comparison to the rest of the class?Select one:a. she did the same on each testb. the math testc. the biology testd. cannot be determined

1 answer:

Anna [14]3 years ago

6 0

Answer:

c. the biology test

Step-by-step explanation:

To answer this problem we need to calculate the z-score of both tests, using the formula:

z = (x - μ) / σ

Where x is Sue's score, μ is the mean, and σ is the standard deviation.

For the <u>biology test</u>, the z-score is:

z = (76 - 70) / 7 = 6/7 = 0.857

For the <u>math test</u>, the z-score is:

z = (76 - 75) / 10 = 1/10 = 0.100

Because the z-score for the biology test is greater than the z-score for the math test, Sue did better in the biology test, in comparison to the rest of the class.

You might be interested in

(HELP ME PLEASE) In the drawing, g > h. Which statement about the volumes of the two cylinders is true?

Aleksandr [31]

Answer:

3rd Option

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Which of the following is the graph of the quadratic function y - x? +10x+16 ?

Assoli18 [71]

Find the vertex:

$\textrm{Use the formula } \frac{-b}{2a} \textrm{ to find the vertex}\\ \\ \frac{-10}{2} = -5\\ \\ \textrm{In order to find the y value plug in -5 back into the function}\\ \\ y=(-5)^2+10(-5)+16\\ y=-9\\\\ \textrm{The vertex of the function is} (-5,-9)$

Determine if the parabola will open up or down:

$a > 0 \\ \\ 1>0$

Since the value of a is positive, the parabola will open upwards

Find the x and y intercepts:

Factor $y = x^2+10x + 16$ to find the x intercepts

(x+8)(x+2)

x + 8 = 0

x = -8

x + 2 = 0

x = -2

x intercepts:

x = -2

x = -8

Find the y intercept

$y = 0^2+10(0) + 16\\\\ y=16$

Hence, C. Graph A best represents the quadratic function $y = x^2+10x + 16$

6 0

3 years ago

RUDIKE [14]

Answer:

$15

Step-by-step explanation:

60x75%

60x0.75

45

60-45=15

5 0

3 years ago

500th term of 24,31,38,45,52

tensa zangetsu [6.8K]

This looks like a simple arithmetic series. Each term is 7 greater than the previous term.

Therefore, the Nth term is 17 + 7N.

For example, the first term is 17 + 7*1 = 17 + 7 = 24.

The second term is 17 + 7*2 = 17 + 14 = 31

etc...

And the 500th term is 17 + 7 * 500 = 17 + 3500 = 3517

3 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex

Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

<u>Given</u>:

$y=\left(x+\sqrt{1+x^2}\right)^m$

<u>First derivative</u>

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}$

<u />

<u /> $\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}$

<u />

$\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}$

<u>Second derivative</u>

<u />

$\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}$

$\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}$

$\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m$

$\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m$

$\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}$

$= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)$

<u>Proof</u>

$(x^2+1)y_2+xy_1-m^2y$

$= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]$

$= \left(x+\sqrt{1+x^2}\right)^m\left[0]$

$= 0$

8 0

1 year ago