If mike reads 50 pages an hour how long would it take him to read 1225 pages

Len [333]

Answer:

A

Step-by-step explanation:

This is exponential decay; the height of the ball is decreasing exponentially with each successive drop. It's not going down at a steady rate. If it was, this would be linear. But gravity doesn't work on things that way. If the ball was thrown up into the air, it would be parabolic; if the ball is dropped, the bounces are exponentially dropping in height. The form of this equation is

$y=a(b)^x$ , or in our case:

$A(n)=a(b)^{n-1}$ , where

a is the initial height of the ball and

b is the decimal amount the bounce decreases each time. For us:

a = 1.5 and

b = .74

Filling in,

$A(n)=1.5(.74)^{n-1}$

If ww want the height of the 6th bounce, n = 6. Filling that into the equation we already wrote for our model:

$A(6)=1.5(.74)^{6-1}$ which of course simplifies to

$A(6)=1.5(.74)^5$ which simplifies to

$A(6)=1.5(.22190066)$

So the height of the ball is that product.

A(6) = .33 cm

A is your answer

3 0

3 years ago

MULTIPLE CHOICE QUESTION<br> ers<br> What is 0.25 as a fraction?

DanielleElmas [232]

$\frac{1}{4}$

Hope it helps

4 0

2 years ago

Read 2 more answers

Determine whether the following is a probability distribution. If not, identify

shutvik [7]

Due to a negative value of P(X = x), it is found that the following is not a probability distribution.

---------------------

A sequence (x, P(X = x)) is a probability distribution if:

The sum of all values of P(X = x) equals 1.
There is no negative values of P(X = x).

In this problem, P(X = 2) is negative, thus, it is not a probability distribution.

A similar problem is given at brainly.com/question/24224213

4 0

2 years ago

A person works 40 hours per week earns 59,70 per year, estimate that persons hourly wage by rounding to the nearest ten thousand

olganol [36]

Answer:

he earns about 120 bucks a week

Step-by-step explanation:

There are 168 hours in a week and, so if the hours in 5970 divided by 50 equals 119.4 (possibly irrelevant) but if he gets paid 5970 per year and works 0.57 hours per day (40 divided by 24) and works about 1.666_ days a week than that would equal 120.85

8 0

3 years ago

At an airport, 79% of recent flights have arrived on time. A sample of 7flights is studied. a.Compute the mean of this probabili

xxTIMURxx [149]

Answer:

a. 5.53

b. 1.078

c. 0.126

d. 0.109

e. 0.549

f. 0.834

g. 0.451

Step-by-step explanation:

The percentage of the flights that arrive on time, P(x) = 79%

The number of flights in the sample, n = 7 flights

a. The mean of the probability distribution, μ = ∑x·P(x)

Therefore, we have; μₓ = n·p

μₓ = 7 × 79/100 = 5.53

b. The standard deviation, σₓ = √(n·p·(1 - p))

∴ σₓ = √(7 × 0.79 × (1 - 0.79)) ≈ 1.078

c. We have;

p = 0.79

q = 1 - p = 1 - 0.79 = 0.21

By binomial probability distribution formula, we have;

The probability of exactly four, P(Exactly 4) = ₇C₄·p⁴·q³

P(Exactly 4) = 35 × 0.79⁴×0.21³ ≈ 0.12625

d. The probability of less than 4 is given as follows;

P(Less than 4) = ₇C₀·p⁰·q⁷ + ₇C₁·p¹·q⁶ + ₇C₂·p²·q⁵ + ₇C₃·p³·q⁴

∴ P(Less than 4) = 1×0.79^0 * 0.21^7 + 7 * 0.79^1 × 0.21^6 + 21*0.79^2*0.29^5+ 85×0.79^3*0.21^4 ≈ 0.109

The probability of less than 4 is ≈ 0.109

e. The probability that more than 5 is given as follows;

P(More than 5) = ₇C₆·p⁶·q¹ + ₇C₇·p⁷·q⁰

7×0.79^6 * 0.21 + 1 * 0.79^7 × 0.21^0 ≈ 0.549

f. The probability that at least 5 of the flight were on time is given as follows;

P(At least 5) = ₇C₅·p⁵·q² + ₇C₆·p⁶·q¹ + ₇C₇·p⁷·q⁰

∴ P(At least 5) = 21×0.79^5 * 0.21^2 + 7×0.79^6 * 0.21 + 1 * 0.79^7 × 0.21^0 ≈ 0.834

g. For the probability that no more than 5 of the flights were on time, e have;

P(At most 5) = 1 - P(More than 5)

∴ P(At most 5) = 1 - 0.549 ≈ 0.451.

6 0

2 years ago