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alina1380 [7]
4 years ago
12

Margaret has 16 yards of fabric and she is making dresses for z doll write the expression for the length of fabric used for each

doll
Mathematics
1 answer:
emmasim [6.3K]4 years ago
7 0
<h3>The expression for the length of fabric used for each doll is \frac{16}{z}\ yards</h3>

<em><u>Solution:</u></em>

Given that,

Margaret has 16 yards of fabric

she is making dresses for z doll

Therefore,

Total length of fabric = 16 yards

Number of dresses = z dolls

Therefore,

\text{ length of fabric used for each doll } = \frac{ \text{ Total length of fabric }} {\text{ z dolls }}\\\\\text{ length of fabric used for each doll } = \frac{16}{z}\ yards

Thus, the expression for the length of fabric used for each doll is \frac{16}{z}\ yards

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Answer:

By definition, the derivative of f(x) is

lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

Let's use the definition for f(x)=\frac{1}{x}

lim_{h\rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{x-(x+h)}{x(x+h)}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{(-1)h}{x^2+xh}}{h}=\\lim_{h\rightarrow 0} \frac{(-1)h}{h(x^2+xh)}=\\lim_{h\rightarrow 0} \frac{-1}{x^2+xh)}=\frac{-1}{x^2+x*0}=\frac{-1}{x^2}

Then, f'(x)=\frac{-1}{x^2}

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4 years ago
5 burgers and 3 fries cost $23.
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4 years ago
Points D, E, and F are on circle C and EF ≅ DF. What is the measure of ∠CDF?
antiseptic1488 [7]
The answer is 

26 degrees 

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6 0
3 years ago
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.<br> Put f(x) = x2 + 3 in y2.
saveliy_v [14]

Answer:

(X2+3)2

Step-by-step explanation:

First, you have to substitute in what f(x) equals for y since f(x) stands for y.

By doing this, the equation turns into (x2+3)2, because x2+3 is y, and 2 because the original equation way y2.

After you substitute, the answer is (x2+3)2.

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3 years ago
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp
olga55 [171]

Answer:

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = \overline{x} = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data​ speeds i.e. x - \overline{x}

x - \overline{x} = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between​ carrier's highest data speed and the mean of all 50 data​ speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

\frac{59.98}{20.03}=2.99

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

z=\frac{x-\overline{x}}{s}

Using the given values, we get:

z=\frac{75.6-15.62}{20.03}=2.99

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly​ high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the  carrier's highest data speed​ is significantly higher.

3 0
3 years ago
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