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3 years ago

What is the exam score for an exam whose z-score is 1.25?

Mathematics

1 answer:

mojhsa [17]3 years ago

4 0

Answer:

Step-by-step explanation:

A z-score of 1.25 means that the actual exam score is 1.25 standard deviations above the mean, and therefore the exam score we are looking for is: mean + 1.25 * SD = 75 + 1.25 * 8 = 85

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What is 41÷0•2 that is the question I have

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Answer:

205

Step-by-step explanation:

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4 years ago

Andrew earns 10 points for every word she spells correctly on a quiz. Which equation can be used to find p, the total number of

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the measures of all of the angles.

irga5000 [103]

Answer:

1. ABC = 60°

2. BCA = 90°

3. CAB = 30°

XVR = WVS = 125°

RVS = 55°

RST = VSU = 110°

RSV = UST= 70°

TUS = 39°

Step-by-step explanation:

According this is a Right Triangle ( BCA = 90°) and the sum angles of a triangle is 180° :

90+5x+9x+6=180

14x=84

x=6

Then we have : ABC = 9(6)+6 =60

CBA = 5(6)=30

b :

Given that the size of the opposite angles to a vertex are equal and the sum of the interior angles of a triangle is 180 degrees, the size of all angles can be easily found.

8 0

2 years ago

Marine biologists have determined that when a shark detectsthe presence of blood in the water, it will swim in the directionin w

siniylev [52]

Solution :

a). The level curves of the function :

$C(x,y) = e^{-(x^2+2y^2)/10^4}$

are actually the curves

$e^{-(x^2+2y^2)/10^4}=k$

where k is a positive constant.

The equation is equivalent to

$x^2+2y^2=K$

$\Rightarrow \frac{x^2}{(\sqrt K)^2}+\frac{y^2}{(\sqrt {K/2})^2}=1, \text{ where}\ K = -10^4 \ln k$

which is a family of ellipses.

We sketch the level curves for K =1,2,3 and 4.

If the shark always swim in the direction of maximum increase of blood concentration, its direction at any point would coincide with the gradient vector.

Then we know the shark's path is perpendicular to the level curves it intersects.

b). We have :

$\triangledown C= \frac{\partial C}{\partial x}i+\frac{\partial C}{\partial y}j$

$\Rightarrow \triangledown C =-\frac{2}{10^4}e^{-(x^2+2y^2)/10^4}(xi+2yj),$ and

$\triangledown C$ points in the direction of most rapid increase in concentration, which means $\triangledown C$ is tangent to the most rapid increase curve.