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Whitepunk [10]
3 years ago
15

Alejandra works at corys coffee shop after school somedays. A customer orders comes to 4.70. The customer hands alejandra a $5 d

ollar bill, so she gives the customer 30 cents in change. Because alejandra loves math , she begins to think about different ways in which she could have given the customer 30 cents in change. Using only quaters, dime, nickels how many ways
Mathematics
1 answer:
vovikov84 [41]3 years ago
7 0

Answer:

5

Step-by-step explanation:

6 nickles,3 dimes ,1 quarter +1 nickel, 2 dimes 2 nickles, 1 dime 4 nickles,

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Consider the one-sided confidence interval expressions for the mean of a normal population. Round your answers to 2 decimal plac
ser-zykov [4K]

Answer: a) 0.84  b) 0.67  c) 1.28

Step-by-step explanation:

Using the standard normal distribution table for z-value , we have

(a) The value of z_{\alpha} would result in a 80% one-sided confidence interval : z_{(1-0.80)}=z_{0.20}=0.8416\approx0.84

(b) The value of z_{\alpha} would result in a 85% one-sided confidence interval : z_{(1-0.85)}=z_{0.25}=0.6744897\approx0.67

(c) The value of z_{\alpha} would result in a 90% one-sided confidence interval : z_{(1-0.90)}=z_{0.10}=1.2815515\approx1.28

6 0
3 years ago
CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is
Rufina [12.5K]

Answer:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

y=\sin(x)\cos(x)

Take the derivative of both sides with respect to x:

\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]

We need to use the product rule:

(uv)'=u'v+uv'

So, differentiate:

y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]

Evaluate:

y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))

Simplify:

y'=\cos^2(x)-\sin^2(x)

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

0=\cos^2(x)-\sin^2(x)

Now, let's solve for x. First, we can use the difference of two squares to obtain:

0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))

Zero Product Property:

0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)

Solve for each case.

Case 1:

0=\cos(x)-\sin(x)

Add sin(x) to both sides:

\cos(x)=\sin(x)

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

0=\cos(x)+\sin(x)

Subtract sine from both sides:

\cos(x)=-\sin(x)

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

And we're done!

Edit: Small Mistake :)

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2 years ago
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Nezavi [6.7K]

You need to explain to us what to do or else nobody will understand what your trying to ask for.

4 0
2 years ago
Use the dot product to find |v| when v = (-2,-1)<br> a. -1<br> b. -3<br> C. Root 5<br> D. Root 3
Firdavs [7]
Answer
Root 5 for sure
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3 years ago
Can someone help me answer this
Firlakuza [10]

Answer:

maby c

Step-by-step explanation:

tell me if this helps

5 0
2 years ago
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