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3 years ago

1) Of the last 6 flights to arrive at Wildgrove Airport, 3 were late.

Mathematics

2 answers:

Oduvanchick [21]3 years ago

5 0

The android to this problem is b.7flights

Lostsunrise [7]3 years ago

4 0

Answer: Jimmy Fallon

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Which expression is equivalent to (4p^-4 q)^-2/ 10pq^-3?

Anit [1.1K]

Answer:

$\dfrac{p^{7} q}{160}$

Step-by-step explanation:

$\dfrac{(4p^-4 q)^-2}{10pq^-3} =$

$= \dfrac{4^{-2}p^{-4\times (-2)} q^{-2}}{10pq^-3}$

$= \dfrac{4^{-2}}{10}p^{8-1} q^{-2-(-3)}$

$= \dfrac{1}{4^2 \times 10}p^{7} q^{-2 + 3}$

$= \dfrac{1}{160}p^{7} q$

$= \dfrac{p^{7} q}{160}$

4 0

3 years ago

Read 2 more answers

Determine whether the graph can represent a normal curve. If it cannot, explain why.

FromTheMoon [43]

Answer:

The answer would be C

Step-by-step explanation:

The answer would be C because It is the most resnible and becaue I took that not to long ago.

Hope this helps

Please give me Brainliest

7 0

3 years ago

The price of a toy usually costs £50 is increased to £65 work out the percentage increase

I am Lyosha [343]

Answer:

Step-by-step explanation:

10% of 50 = 5

5% of 50 = 2.5

to get to 65 u do 5+5 = 10 so so far that is 20% but 20% only gets u to 60 so u add another 10% to reach 65

u have used 30% so the answer to yr question is 30%

<h2><em>does this help comment if / if not </em></h2>

8 0

3 years ago

What is the range for the graph below?

mario62 [17]

Answer:

All real numbers except -3

Step-by-step explanation:

Range refers to numbers on the y-axis.

Domain refers to numbers on the x-axis.

Both of those lines will never reach -3, just come extremely close to it.

6 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.

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