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azamat
3 years ago
6

The fastest a human has ever run is 27 miles per hour. How many miles per minute did the human run?

Mathematics
1 answer:
erastovalidia [21]3 years ago
4 0
27 miles per hour

1 hour = 60 minutes 

27 miles per 60 minutes

27/60 miles per 1 minute

9/20 mile per minute
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If the angles 2xº and 4yº<br> are<br> complementary, find the value of x + 2y.
ivolga24 [154]

Answer:

Solution given:

the angles 2xº and 4yºare complementary

so

2x+4y=90°

dividing both side by 2we get

x+2y=45°

<u>the value of x + 2y</u><u> </u><u>i</u><u>s</u><u> </u><u>4</u><u>5</u><u>°</u><u>.</u>

3 0
3 years ago
A part selected for testing is equally likely to have been produced on any one of six cutting tools.
JulijaS [17]

Answer:

a)S= (1,2,3,4,5,6)

b) 1/6 or 16.67%

c) 1/3 or 33.33%

d) 5/6 or 83.33%

Step-by-step explanation:

a) Sample space is the list of all the possible sample in this its either tool 1,2,3,4,5 or 6

b) probability of selecting from tool 1 is equally likely to picking from any cutting tool therefore it

1/6 or 16.7%

c) probability of selecting from any tool is 16.7% so probability of selecting from 3 or 5 = 16.67% + 16.67% = 33.3%

d) if the part is not from tool 4 then its from tool 1, or tool 2 or tool 3 or tool 5 or tool 6 therefore its

1/6+1/6+1/6+1/6+1/6 = 5/6 or 8.33%

8 0
3 years ago
The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
Lemur [1.5K]

Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

4 0
3 years ago
What is the rate of change of the function? <br> -3<br> -1/3 <br> 1/3<br> 3
sweet-ann [11.9K]

Answer: the rate change is 1/3

Step-by-step explanation: because its going up 1/3 not down.


3 0
3 years ago
Read 2 more answers
I will give brainliest to people who answer ALL 5 questions.
amid [387]
Here ya go. Hope this helps. Have a great day

8 0
2 years ago
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