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Zigmanuir [339]
3 years ago
12

A) A box contains 50 diodes of which 10 are known to be bad. A diode is selected at random. What...

Mathematics
1 answer:
lubasha [3.4K]3 years ago
4 0
A) is not likely that it's bad B)it would be more likely C)more probable
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A differential equation that is a function of y only
Pachacha [2.7K]
ANSWER
<span>will produce a slope field with rows of parallel tangents 
</span>
EXPLANATION
A differential equation that is a function of y only does not depend on x, i.e. of the form dy/dx = f(y), where f is a function.

Therefore, the slope field will be independent of the horizontal position. This means that there will be rows where the slopes will be of the same rise/run ratio. When two slopes are the same, they are said to be parallel.
7 1
3 years ago
Can somebody help me on this one?
Wittaler [7]

Answer:

I pretty sure the answer is Nervous

5 0
3 years ago
Read 2 more answers
How does replacing f(x) with f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative) affect
ivann1987 [24]
F(x) + k - Moves the graph k units up.

k f(x)   stretches the graph parallel to y-axis by a facor k

f (kx) stretches  the  graph by a factor 1/k parallel to x-axis

f(x + k)  moves the graph   3 units to the left.

For k negative the first one moves it k units down

for second transform negative does same transfoormation but also reflects the graph in the x axis

For the third transform negative k :- same as above but also reflects in y axis

4th transform -  negative k moves graph k units to the right


4 0
3 years ago
Find the exact value of cos theta​, given that sin thetaequalsStartFraction 15 Over 17 EndFraction and theta is in quadrant II.
vova2212 [387]

Answer:

cos \theta = -\frac{8}{17}

Step-by-step explanation:

For this case we know that:

sin \theta = \frac{15}{17}

And we want to find the value for cos \theta, so then we can use the following basic identity:

cos^2 \theta + sin^2 \theta =1

And if we solve for cos \theta we got:

cos^2 \theta = 1- sin^2 \theta

cos \theta =\pm \sqrt{1-sin^2 \theta}

And if we replace the value given we got:

cos \theta =\pm \sqrt{1- (\frac{15}{17})^2}=\sqrt{\frac{64}{289}}=\frac{\sqrt{64}}{\sqrt{289}}=\frac{8}{17}

For our case we know that the angle is on the II quadrant, and on this quadrant we know that the sine is positive but the cosine is negative so then the correct answer for this case would be:

cos \theta = -\frac{8}{17}

5 0
3 years ago
There are 48 students in the 6th-grade band. If this represents 30% of the students in 6th grade, how many students are in the 6
Artemon [7]

Answer:

160

Step-by-step explanation:

had the question before

4 0
2 years ago
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