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2 years ago

Please help 13 MATH PROBLEM

Mathematics

2 answers:

padilas [110]2 years ago

6 0

Answer: The answer is 431/120 then you Simplify it and that would be

3 71/120 so the answer is F

Step-by-step explanation:

Hope this helps ya! Have a good day! :)

Alexxx [7]2 years ago

6 0

I’m pretty sure the answer is g

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A particle is moving on a straight line in such a way that its velocity v is given by v ( t ) = 2 t + 1 for 0 ≤ t ≤ 5 where t is

Lady_Fox [76]

Answer:

The total distance traveled by the particle is S = 30.

Step-by-step explanation:

Given that velocity,

v(t) = 2t + 1

To find the total distance travel, we integrate the velocity function, v(t), to obtain the distance function s(t), and evaluate the resulting distance at the interval given. That is at t = 0 to t = 5.

Integrating v(t) with respect to t, we have

s(t) = t² + t + C.

At t = 5

s(5) = 5² + 5 + C

= 25 + 5 + C

= 30 + C

At t = 0

S(0) = 0 + 0 + C

= C

The required distance is now

S(5) - S(0)

= 30 + C - C

= 30.

8 0

2 years ago

Read 2 more answers

30 POINTS + BRAINLIEST TO FIRST CORRECT ANSWER!

Vinvika [58]

Answer:

The length will be multiplied by 2.

Step-by-step explanation:

By looking at the first and third ratio, we know that the no. of birck have been multiplied by 2 and the length has also been multiplied by 2 .

So when the number of bricks is multiplied by 2, the length is multiplied by 2

7 0

2 years ago

Read 2 more answers

A yellow ribbon is 56 centimeters long. It is twice as long as a green ribbon. A brown ribbon is 4 times as long as the green ri

Shkiper50 [21]

56 dived by 2 equals 28. 28=green ribbon. Brown ribbon is 4 times as long. 28 times 4 equals 112. The brown ribbon is 112 cm long.

6 0

2 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

2 years ago

Compute the multifactor productivity measure for each of the weeks shown for production of chocolate bars. What do the productiv

liq [111]

Answer:

Week 1 =3.03

Week 2 =2.99

Week 3=2.9

Week 4=2.83

Step-by-step explanation:

Computation for the multifactor productivity measure for each of the weeks shown for production of chocolate bars.

Using this formula

Week Multifactor productivity measure = (Output / Total Cost)

Where:

Output represent Quantity produced

Total cost represent Labor cost + Material cost + Overhead

Let plug in the formula

Week 1= 30,000/(40*12*6)+[(40*12*6)1.5]+(450*6)

Week 1 =30,000/(2,880)+(2,880*1.5)+2,700

Week 1 =30,000/(2,880+4,320+2,700)

Week 1=30,000/9,900

Week 1 =3.03

Week 2= 33,600/(40*12*7)+[(40*12*7)1.5]+(470*6)

Week 2 =33,600/(3,360)+(3,360*1.5)+2,820

Week 2=33,600/(3,360+5,040+2,820)

Week 2=33,600/11,220

Week 2 =2.99

Week 3= 32,200/(40*12*7)+[(40*12*7)1.5]+(460*6)

Week 3 =32,200/(3,360)+(3,360*1.5)+2,760

Week 3=32,200/(3,360+5,040+2,760)

Week 3=32,200/11,160

Week 3=2.9

Week 4= 35,400/(40*12*8)+[(40*12*8)1.5]+(480*6)

Week 4 =35,400/(3,840)+(3,840*1.5)+2,880

Week 4=35,400/(3,840+5,760+2,880)

Week 4=35,400/12,480

Week 4=2.83

Therefore the multifactor productivity measure for each of the weeks shown for production of chocolate bars is :

Week 1 =3.03

Week 2 =2.99

Week 3=2.9

Week 4=2.83

4 0

2 years ago