All categories

3 years ago

G use part 1 of the fundamental theorem of calculus to find the derivative of the function. sqrt(4+7t)

Mathematics

1 answer:

uysha [10]3 years ago

6 0

Let y = √4+7t
then u= 4+7t
y=√u = u^½

du/dt= 7
dy/du = ½U^-½

dy/dt = du/dt • dy/du
= 7×½U^-½
= 7/2√U
= 7 / (2{√4+7t})

You might be interested in

etta bought a calculator for $15. Glenn found the same model for $9.79. How much more did Etta pay than Glenn

ololo11 [35]

5.21.

To find this, give both numbers decimal points before setting up a subtraction equation as so:

15.00 - 9.79

Then proceed to solve to get the final answer of 5.21.

Hope this helps!

6 0

3 years ago

Read 2 more answers

A dice is rolled 2 times. What is the probability of showing a 5 on every roll Round to 4 decimal places

AlexFokin [52]

Answer:

0.0278

Step-by-step explanation:

(1/6)(1/6) = 1/36 = 0.0278

3 0

3 years ago

Priscilla can make 4 bracelets in 17 minutes. At this rate, how many bracelets can she make in 34 minutes?

andrew11 [14]

4 bracelets/17 minutes = x bracelets/34 minutes
cross multiply: 4 · 34 = 17 · x
Divide both sides by 17: (4 · 34)/17 = x
x = 8

Answer: 8 bracelets

6 0

3 years ago

During his hike milt drank 1 liter of water and 1 liter of sports drink how many milliliters of liquid did he drink

a_sh-v [17]

In total, he drank 2 liters.

In millimeters, this is:
2,000 milliliters.

1 liter = 1,000 milliliters
2 liters = 1,000×2
= 2,000 millimetres

3 0

3 years ago

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

3 years ago