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kirill115 [55]
3 years ago
6

Topic 1 performance assessment pg. 53 and 54

Mathematics
2 answers:
777dan777 [17]3 years ago
7 0
What are u talking about? i don't have an performance assessment of any kind 
Dmitrij [34]3 years ago
7 0
 <span>Mrsfranny911gmailcom, please attach a document or explain what the question is</span>
You might be interested in
2/3 + (-5/7)<br><br><br><br> .........
dmitriy555 [2]

Answer:

-1/21

Step-by-step explanation:

2/3 + -5/7

Get a common denominator of 21

2/3 *7/7 = 14/21

-5/7*3/3 = -15/21

14/21 - 15/21 = -1/21

4 0
4 years ago
Read 2 more answers
The vertices of a triangle are P(−3, −4), Q(3, 4), and R(−6, −3). Name the vertices of Rx = 0 (PQR).
Katyanochek1 [597]
The correct answer is <span>A) P'(3, −4), Q'(−3, 4), R'(6, −3)</span>

Rx = 0 indicates a reflection over the y-axis. 

The rule for such a transformation is:
(x, y) --> (-x, y)
which means that the x-coordinate changes sign and the y-coordinate stays the same.

Therefore:
P<span>(-3, -4) --> P'(3, -4)
Q(3, 4) --> Q'(-3, 4)
R(-6, -3)</span> --> R'(6, -3)

These points are those in option A).
6 0
3 years ago
Rob is saving to buy a new MP3 player. For every $15 he earns babysitting, he saves $9. On Saturday, Rob earned $60 babysitting.
alexandr402 [8]
$60 ÷ $15 = 4
$9 × 4 = $36
he saved $36 when he earned $60
8 0
3 years ago
Consider f and c below. f(x, y, z) = (y2z + 2xz2)i + 2xyzj + (xy2 + 2x2z)k,
qaws [65]
\dfrac{\partial f}{\partial x}=y^2z+2xz^2
f(x,y,z)=xy^2z+x^2z^2+g(y,z)

\dfrac{\partial f}{\partial y}=2xyz+\dfrac{\partial g}{\partial y}=2xyz
\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)

\dfrac{\partial f}{\partial z}=xy^2+2x^2z+\dfrac{\mathrm dh}{\mathrm dz}=xy^2+2x^2z
\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=0

\implies f(x,y,z)=xy^2z+x^2z^2+C
8 0
4 years ago
The average student-loan debt is reported to be $25,235. A student believes that the student-loan debt is higher in her area. Sh
yulyashka [42]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 25235

For the alternative hypothesis,

µ > 25235

This is a right tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 100,

Degrees of freedom, df = n - 1 = 100 - 1 = 99

t = (x - µ)/(s/√n)

Where

x = sample mean = 27524

µ = population mean = 25235

s = samples standard deviation = 6000

t = (27524 - 25235)/(6000/√100) = 3.815

We would determine the p value using the t test calculator. It becomes

p = 0.000119

Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.

6 0
3 years ago
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