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fenix001 [56]
3 years ago
10

Please help me I been stuck for 30m can’t figure it out

Mathematics
1 answer:
Step2247 [10]3 years ago
5 0
The graph is at a slope:1 with a y-intercept:-4

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rusak2 [61]
Answer:
the amount of books grant has left

explanation:
if you take away b books, he is left with 29-b books
3 0
2 years ago
The regular price for a sweater is $48. The store is having buy one get one 1/2 off sale. If you buy 2 sweaters for that price w
IRISSAK [1]
25%


If you are buying 2 shirts for the same price, you have to calculate the cost of the shirt at half price first (1 piece of the pie) and then add it the original price (2 equal pieces of the pie since you just divided that to the price of the second shirt). That would be considered 3 out of 4 pieces off the pie or 75%. Your savings would be 25%.
4 0
3 years ago
Read 2 more answers
A spinner has equal regions numbered 1 through 20. What is the probability that the spinner will stop on an odd number or a mult
klio [65]

Answer:

<em>Probability that the spinner will stop on an odd number or a multiple of 5 is </em><em>0.6</em>

Step-by-step explanation:

Probability = \frac{Required outcomes}{Total possible outcomes}

We are given the equal regions numbered from 1 through 20 which means that our total possible outcomes are 20

<em>Total possible outcomes: 20</em>


<em>Outcomes that spinner will stop on an odd number, n(Odd): 10</em>

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Probability of spinner stoping on Odd number:

P(Odd) = \frac{n(Odd)}{Total} = \frac{10}{20} = \frac{1}{2} = 0.5


Outcomes that spinner will stop on a multiple of 5, n(5): 4

5, 10, 15, 20

Probability of spinner stoping on multiple of 5:

P(5) = \frac{n(5)}{Total} = \frac{4}{20} = \frac{1}{5} = 0.2

Odd numbers which are a multiple of 5 are: 5 and 15

So,

P(Odd and 5) = \frac{2}{20}=\frac{1}{10}=0.1

Thus Probability of spinner stopping at odd number or a multiple of 5 becomes:

P(Odd or 5) = P(Odd) + P(5) - P(Odd and 5) = 0.5 + 0.2 - 0.1 = 0.6

7 0
3 years ago
For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains
AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

f(x) = (1-p)^{x}p

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

\mu = \frac{1-p}{p}

The standard deviation of the geometric distribution is given by the following formula:

\sigma = \sqrt{\frac{1-p}{p^{2}}

In this problem, we have that:

p = 0.383

So

\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61

\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05

(a) What is the probability that a drought lasts exactly 3 intervals?

This is f(3)

f(x) = (1-p)^{x}p

f(3) = (1-0.383)^{3}*(0.383)

f(3) = 0.09

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is P = f(0) + f(1) + f(2) + f(3)

f(x) = (1-p)^{x}p

f(0) = (1-0.383)^{0}*(0.383) = 0.383

f(1) = (1-0.383)^{1}*(0.383) = 0.236

f(2) = (1-0.383)^{2}*(0.383) = 0.146

Previously in this exercise, we found that f(3) = 0.09

So

P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4).

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

P(X \leq 3) + P(X \geq 4) = 1

0.855 + P(X \geq 4) = 1

P(X \geq 4) = 0.145

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0
3 years ago
SUPER EASY FOR POINTS!!!!solve: x-3=3x-2
Savatey [412]

Answer:

Polynomial equation solver

x-3=3x-2

Standard form:

−2x − 1 = 0

Factorization:

−(2x + 1) = 0

Solutions:

x = −1

2

= -0.5

3 0
3 years ago
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