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alexandr1967 [171]

3 years ago

5

A juggler is performing her act using several balls. She throws the balls up at an initial height of 4 feet, with a speed of 15

feet per second. If the juggler doesn't catch one of the balls, about how long does it take the ball to hit the floor?
Hint: Use H(t) = −16t2 + vt + s.

2 answers:

Vikki [24]3 years ago

6 0

I think the answer is 1.15 seconds. Thats just a guess..... So yeah sorry if wrong!! BYYYYYYYYEEEE :Letter b I think IDKKK

jonny [76]3 years ago

5 0

I did this question and got it right its 1.15 seconds

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So what exactly is a,b,c

Gala2k [10]

If you’re talking about answers a would be the option first b would be 2nd and c would be 3rd

4 0

2 years ago

A bicycle wheel has a circumference of 62 inches. Find the diameter of the wheel. If necessary, round to the tenths place. A. 39

san4es73 [151]

Answer:

The answer to your question is diameter = 19.7 in

Step-by-step explanation:

Data

Circumference = 62 in

diameter = ?

Process

1.- Write the formula of Circumference, remember that circumference = perimeter.

Perimeter = 2πr

2.- Equal the equation to the value given

2πr = 62

-Solve it for r

r = 62/2(3.14)

-Result

r = 62/6.28

r = 9.87 in

3.- Find the diameter

diameter = 2r

diameter = 2(9.87)

diameter = 19.7 in

4 0

3 years ago

Read 2 more answers

randy earns $14.50 per hour for the first 40 hours worked in a week and $21.75 per hour for hours over 40. Last week Randy earne

Nutka1998 [239]

Answer:

4 hours

Step-by-step explanation:

We know the following:

Total amount Randy earned: $667

Pay for first 40 hours: $14.50 per hour

Pay after 40 hours: $21.75 per hour

We need to find the total extra hours: $x$

If the total number of hours worked are, $T$ , then: $x + 40 = T$

Total pay received for the first 40 hours: $40 \times 14.50 = 580$

If out of $667, Randy earns $580 from the first 40 hours, the left over are earned by extra hours.

$667 - $580 = $87 ------- (a)

Total pay received for working after 40 hours: $21.75x$

We know from eq (a) that total pay is 87, hence we can find the number of hours using this.

$21.75x = 87\\\\x = \frac{87}{21.75}\\\\x = 4$

Randy worked 4 extra hours.

6 0

2 years ago

You have a dance class every three days. Today is Monday, and you have dance class. In how many more days will you have dance cl

Dafna11 [192]

7 days in a week, 3 days a rotation.

multiples of 7: 7, 14, 21, 28

multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24

the LCM is 21, because it is the smallest multiple of both numbers. there will be 21 more days until you take dance class on Mondays

8 0

3 years ago

Read 2 more answers

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi

svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

3% of the components are identified as defective

This means that $p = 0.03$

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is $P(X = 0)$ when $n = 90$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645$

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so $P(X \leq 2)$ when $n = 102$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447$

$P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411$

$P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062$

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so $P(X \leq 5)$ when $n = 105$

Then

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408$

$P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326$

$P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133$

$P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265$

$P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786$

$P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034$

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

3 0

3 years ago