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motikmotik
3 years ago
7

Danica is making a rectangular isle runner which has a length that is 8 inches more than four times the width. Create a function

, A(w), for the area of the isle runner, where w represents the width in inches.
Mathematics
1 answer:
xxTIMURxx [149]3 years ago
4 0

Answer:

Step-by-step explanation:

L=4W+8

A=LW, using L from above makes this

A(W)=(4W+8)W

A(W)=4W^2+8W

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Z* for 90% confidence is ________. (hint: use z-table c, or software.) answer to three decimal places x.xxx
Aloiza [94]

Answer:

We have to find the z critical value at 100% - 90% = 10% level of significance.

Since the normal distribution has two tails, we therefore, need to divide the significance level into two tails with each tail having area of 5%.

Now using the standard normal table, the z critical is:

z=\pm1.645



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Whats the square root of 125
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The square root of 125 is 11.18033988749895

Step-by-step explanation:


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the owner of an electorinics store recived a shipment of mp3 players at a cost of 64.25 each. If he sells the mp3 players for $9
MissTica

Answer:

he would make a profit of $35.74

the percent of markup is 56% (55.62 to be exact)

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3 years ago
The numbers of girls in a second grade class in 4 less than the number of boys. Let b represent the number of boys in the class,
liubo4ka [24]

Answer:

Step-by-step explanation:

But an B

6 0
3 years ago
How do you do these problems?
Angelina_Jolie [31]

Answer:

18e⁶

⁵/₁₂₈

Step-by-step explanation:

Rₙ(x) = f⁽ⁿ⁺¹⁾(c) / (n+1)! (x − a)ⁿ⁺¹, and a < c < x.

f(x) = eˣ, a = 0, and n = 1.  Thus R₁ is:

R₁(x) = f"(z)/2! x²

R₁(x) = eᶻ/2 x²

|R₁| is a maximum when |f"(z)| is a maximum.  On the domain 0 < z < 6, that maximum is e⁶.  At x = 6, the upper bound of |R₁| is:

|R₁| = 18e⁶

This time, f(x) = 1 / √(1 + x) = (1 + x)^-½.  a = 0, and n = 2.

R₂(x) = f⁽³⁾(z)/3! x³

Find f⁽³⁾(x):

f'(x) = -½ (1 + x)^-³/₂

f"(x) = ¾ (1 + x)^-⁵/₂

f⁽³⁾(x) = -¹⁵/₈ (1 + x)^-⁷/₂

On the domain -½ < z < 0, |f⁽³⁾(z)| is a maximum at z = 0.

|f⁽³⁾(z)| = ¹⁵/₈

Therefore, at x = -½, the upper bound of R₂ is:

|R₂| = (¹⁵/₈)/6 |(-½)³|

|R₂| = ⁵/₁₂₈

8 0
3 years ago
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