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2 years ago

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An object is launched straight into the air. The projectile motion of the object can be modeled using h(t) = 96t – 16t2, where t

is the time since launch and h(t) is the height in feet of the projectile after time t in seconds.

1 answer:

ratelena [41]2 years ago

6 0

Answer with Step-by-step explanation:

We are given that height of projectile after t seconds is given by

$h(t)=96t-16t^2$

a.h(t)=144 ft

$144=96t-16t^2$

$16t^2-96t+144=0$

$t^2-6t+9=0$

$t^2-3t-3t+9=0$

$t(t-3)-3(t-3)=0$

$(t-3)(t-3)=0$

t-3=0

t=3

After 3 s, the height of the project will be 144 feet in the air.

b.h(t)=0

$96t-16t^2=0$

$16t(6-t)=0$

$16t=0\implies t=0$

$6-t=0\implies t=6$

At t=0, the initial position of projectile

At t=6 s , the projectile will hit the ground.

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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is $\{1,2,3,4\}$ .

a) We need to find a relation R reflexive and transitive that contain the relation $R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}$

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

1R4 and 4R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
4R1 and 1R2, then 4 must be related with 2.

Therefore $\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\}$ is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

since 1R2, then 2 must be related with 1.
since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

1R2 and 2R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
2R1 and 1R4, then 2 must be related with 4.
4R1 and 1R2, then 4 must be related with 2.
2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

1 must be related with 1,

2 must be related with 2,

3 must be related with 3,

4 must be related with 4

For be symmetric

since 1R2, 2 must be related with 1,
since 1R4, 4 must be related with 1.

For be transitive

Since 4R1 and 1R2, 4 must be related with 2,
since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

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1 year ago

Montel collected 26 leaves for his science project. he wants to put 6 leaves on a page. which shows how many pages montel will n

MissTica

You divide 26 by 6, and you get 4 1/3, so I guess Montel will need 5 pages.

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2 years ago

Read 2 more answers

To increase sales, a local donut shop began putting an extra donut in some of the boxes. Customers are unaware of which boxes ha

Bingel [31]

Answer:

Step-by-step explanation:

Hello!

Part A

First, determine your study variable:

X: Number of boxes with an extra donut in a sample of eight.

To see if the variable has a Binomial distribution you have to check if the binomial criteria are met:

1. The number of observation of the trial is fixed (In this case n = 8, the boxes each customer bought make the sample)

2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial (In this case, the amount of donuts in one box does not affect on the probability of the next box having an extra donut)

3. The probability of success in the same from one trial to another (In this case our "success" will be that the box has an extra donut, according to the owners claim that is 1/7; ρ=0,14)

So X≈ Bi (n;ρ)

Where n represents the sample (n=8) and ρ is the probability of success (ρ=0.14)

Part B

The mean of the binomial distribution is E(X)= nρ

E(X)= 8 * 0.14= 1.12

The mean of the distribution is also called the expected value. You'd expect that 1.12 boxes have an extra donut.

The variance of the binomial distribution is V(X)= nρ(1 - ρ)

V(X)= 8*0.14*(1 - 0.14)= 0.9632

Its square root is the standard deviation

√V(X)= 0.98

The standard deviation is a measure of dispersion, it indicates how much the values of the distribution of the central value are separated. In this case, 0.98 indicates that the distribution of the number of boxes with an extra donut is far from the expected value.

Part C

Two of the eight customers buy a box with an extra donut, symbolically:

P(X=2) = P(X≤2) - P(X≤1)= 0.91 - 0.68 = 0.22

There is a 22% chance that two customers bought a box with an extra donut.

Compute:

P(X≥2)= 1 - P(X<2)= 1 - P(X≤1)= 1 - 0.68= 0.32

There is a 32% chance that two or more customers bought a box with an extra donnut.

I hope it helps!

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1 year ago

A psychologist is collecting data on the time it takes to learn a certain task. For 50 randomly selected adult subjects, the sam

tatuchka [14]

Answer: $(15.263,\ 17.537)$

Step-by-step explanation:

According to the given information, we have

Sample size : n= 50

$\overline{x}=16.40$

$s=4.00$

Since population standard deviation is unknown, so we use t-test.

Critical value for 95 percent confidence interval :

$t_{n-1,\alpha/2}=t_{49, 0.025}= 2.009575\approx2.010$

Confidence interval : $\overline{x}\pm t_{n-1, \alpha/2}\dfrac{s}{\sqrt{n}}$

$16.40\pm (2.010)\dfrac{4}{\sqrt{50}}\\\\=16.40\pm1.13702770415\\\\=16.40\pm1.1370\\\\=(16.40-1.1370,\ 16.40+1.1370)\\\\=(15.263,\ 17.537)$

Required 95% confidence interval : $(15.263,\ 17.537)$

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2 years ago

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Answer:

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Step-by-step explanation:

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$y = 3\cdot x^{2}-3\cdot x +2$

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Now, this system is now represented by means of a graphing tool and whose outcome is attached below. There are two solutions: (0, 2) and (1, 2)

ii) Let be $y=3\cdot x^{2}-3\cdot x -1$ , if $3\cdot x^{2}-3\cdot x +2 = x+1$ , which is equivalent to the following system of equations:

$y = 3\cdot x^{2}-3\cdot x +2$

$y = x+1$

Now, this system is now represented by means of a graphing tool and whose outcome is attached below. There are two solutions: (0.333, 1.333) and (1, 2)

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