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Papessa [141]
3 years ago
7

Cathie is making a dress with a sash. The sash needs to be 72.8 inches long, but Cathie's measuring tape is only in centimeters.

Which of the following shows the number of centimeters, with the correct number of significant digits, that Cathie needs to measure in order for the sash to be 72.8 inches long? (1 inch = 2.54 centimeters)
Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
4 0
<span>The sash needs to be 184.912 cm. to equal 72.8 in.</span>
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Find
Westkost [7]

Answer:

dy/dx = -b/a cot α

Step-by-step explanation:

x² / a² + y² / b² = 1

Take derivative with respect to x.

2x / a² + 2y / b² dy/dx = 0

2y / b² dy/dx = -2x / a²

dy/dx = -b²x / (a²y)

Substitute:

dy/dx = -b²a cos α / (a²b sin α)

dy/dx = -b cos α / (a sin α)

dy/dx = -b/a cot α

8 0
3 years ago
An airline claims that the proportion of no-shows for passengers who booked on its flights is less than 0.06. In a random sample
Genrish500 [490]

Step 1:

In a sample of 380 randomly selected reservations, 19 were no-shows.

Step 2:

Proportion of no shows p<0.06.

Step 3:

Test Value

z(19/380)=0.05

Step 4:

Test statistics

a) 0.05-1.124=-1.074

b) 0.05-(-1.943) = 0.05+1.943=1.993

c)0.05-(-0.821)=0.05+0.821=0.871

d)0.05 - 0.222 = - 0.172

e)0.05 -(-1.571) = 0.05+1.571 = 1.621

The above data clearly mentions the test statistics associated with the given samples.

7 0
3 years ago
Read 2 more answers
What is the factored for of x^2-x-2?
Olenka [21]
B because it makes sense
4 0
3 years ago
Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua
goblinko [34]
<h2>Hello!</h2>

The answer is:  There is a total of 5.797 gallons pumped during the given period.

<h2>Why?</h2>

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

D(t)=\frac{5t}{1+3t}

So, the integral will be:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dx

So, integrating we have:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx

Performing a change of variable, we have:

1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}

Then, substituting, we have:

\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du

\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )

\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du

\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]

Reverting the change of variable, we have:

\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]

Then, evaluating we have:

\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0
3 years ago
in one hour thirty two cars pass through a particular intersection at the same rate how long would it take for 96 cars to pass t
USPshnik [31]

Answer:

72 hours

Step-by-step explanation:

1hour30minutes times 48 cars (since TWO pass every 1:30, not one)

equals 72 hours, or three days

3 0
3 years ago
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